Examkin

Fundamental Trigonometric Identities

19 soru

Tüm türler19Çoktan seçmeli9Sayısal10
Soru 1Soru →

For an angle θ\theta in the interval π2<θ<π\frac{\pi}{2} < \theta < \pi, the value of cos⁡θ=−35\cos \theta = -\frac{3}{5}. What is the value of sin⁡θ+cos⁡θ\sin \theta + \cos \theta?

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Cevap: 15\frac{1}{5}

Cevap

one-fifth
To evaluate sin⁡θ+cos⁡θ\sin \theta + \cos \theta, we first find the value of sin⁡θ\sin \theta. We substitute cos⁡θ=−35\cos \theta = -\frac{3}{5} into the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, which gives sin⁡2θ+925=1\sin^2 \theta + \frac{9}{25} = 1. Solving for sin⁡2θ\sin^2 \theta yields 1625\frac{16}{25}. Because the angle θ\theta is constrained to the second quadrant (π2<θ<π\frac{\pi}{2} < \theta < \pi), its sine value must be positive, which means sin⁡θ=45\sin \theta = \frac{4}{5}. Adding the values together, we get 45+(−35)=15\frac{4}{5} + \left(-\frac{3}{5}\right) = \frac{1}{5}.

Adım Adım Çözüm

1
Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to find the magnitude of the sine function.
sin⁡2θ=1−(−35)2=1−925=1625\sin^2 \theta = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}
The Pythagorean identity relates sine and cosine for any angle.
2
Determine the correct sign of sin⁡θ\sin \theta based on the given quadrant interval.
Since π2<θ<π\frac{\pi}{2} < \theta < \pi, the angle θ\theta lies in Quadrant II, where the sine function is positive. Thus, sin⁡θ=1625=45\sin \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}.
The trigonometric function values are positive or negative depending on the quadrant on the unit circle.
3
Compute the sum of sin⁡θ\sin \theta and cos⁡θ\cos \theta.
sin⁡θ+cos⁡θ=45+(−35)=15\sin \theta + \cos \theta = \frac{4}{5} + \left(-\frac{3}{5}\right) = \frac{1}{5}
This gives the final value of the requested expression.

Anahtar Kavram

Using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 and quadrant rules to calculate trigonometric values.

Alternatif Yöntem

We can sketch a reference right triangle in Quadrant II. Since cos⁡θ=adjacenthypotenuse=−35\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = -\frac{3}{5}, we assign the adjacent side a length of −3-3 along the x-axis and the hypotenuse a length of 55. By the Pythagorean theorem, the opposite vertical side is 52−(−3)2=4\sqrt{5^2 - (-3)^2} = 4. Since the vertical side is in Quadrant II, it is positive. This makes sin⁡θ=oppositehypotenuse=45\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}. Evaluating the sum yields 45+(−35)=15\frac{4}{5} + \left(-\frac{3}{5}\right) = \frac{1}{5}.
Tahmini Süre:45s
Soru 2Soru →

For an angle θ\theta such that π<θ<3π2\pi < \theta < \frac{3\pi}{2}, the value of cos⁡θ=−0.8\cos\theta = -0.8. What is the value of 10sin⁡θ10\sin\theta?

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Cevap: -6

Cevap

The value of 10sin⁡θ10\sin\theta is −6-6.
The correct answer is −6-6. Substituting cos⁡θ=−0.8\cos\theta = -0.8 into the fundamental Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 yields sin⁡2θ+0.64=1\sin^2\theta + 0.64 = 1, which simplifies to sin⁡2θ=0.36\sin^2\theta = 0.36. Since the angle θ\theta lies in the interval π<θ<3π2\pi < \theta < \frac{3\pi}{2} (Quadrant III), its sine value must be negative. Thus, sin⁡θ=−0.6\sin\theta = -0.6. Multiplying this value by 10 gives 10sin⁡θ=−610\sin\theta = -6.

Adım Adım Çözüm

1
Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 with the given cosine value cos⁡θ=−0.8\cos\theta = -0.8.
sin⁡2θ+(−0.8)2=1⇒sin⁡2θ+0.64=1⇒sin⁡2θ=0.36\sin^2\theta + (-0.8)^2 = 1 \Rightarrow \sin^2\theta + 0.64 = 1 \Rightarrow \sin^2\theta = 0.36
The Pythagorean identity relates the sine and cosine values of any angle.
2
Take the square root of both sides, selecting the correct sign based on the quadrant constraint π<θ<3π2\pi < \theta < \frac{3\pi}{2}.
Since the angle lies in Quadrant III, the sine function must be negative. Thus, sin⁡θ=−0.36=−0.6\sin\theta = -\sqrt{0.36} = -0.6.
In the third quadrant, y-coordinates on the unit circle are negative, meaning sin⁡θ\sin\theta must be negative.
3
Multiply sin⁡θ\sin\theta by 10 to get the final requested value.
10sin⁡θ=10×(−0.6)=−610\sin\theta = 10 \times (-0.6) = -6
This calculation yields the final answer requested by the problem.

Anahtar Kavram

Applying the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 and choosing the correct sign based on the angle's quadrant.
Soru 3Soru →

An angle θ\theta lies in the second quadrant and satisfies sin⁡θ=513\sin\theta = \frac{5}{13}. What is the value of 12tan⁡θ12\tan\theta?

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Cevap: -5

Cevap

The value of 12tan⁡θ12\tan\theta is -5.
Applying the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 with sin⁡θ=513\sin\theta = \frac{5}{13} yields cos⁡2θ=144169\cos^2\theta = \frac{144}{169}. Since the angle θ\theta lies in the second quadrant, its cosine is negative, meaning cos⁡θ=−1213\cos\theta = -\frac{12}{13}. Using the quotient identity tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}, we find tan⁡θ=−512\tan\theta = -\frac{5}{12}. Multiplying by 12 gives the correct value of -5.

Adım Adım Çözüm

1
Use the Pythagorean identity to find the magnitude of cos⁡θ\cos\theta.
cos⁡2θ=144169\cos^2\theta = \frac{144}{169}
The Pythagorean identity states that sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. Substituting sin⁡θ=513\sin\theta = \frac{5}{13} gives cos⁡2θ=1−(513)2\cos^2\theta = 1 - \left(\frac{5}{13}\right)^2.
2
Determine the value of cos⁡θ\cos\theta by applying the quadrant sign rule.
cos⁡θ=−1213\cos\theta = -\frac{12}{13}
Since θ\theta is in the second quadrant, the cosine of θ\theta must be negative.
3
Calculate tan⁡θ\tan\theta using the quotient identity.
tan⁡θ=−512\tan\theta = -\frac{5}{12}
The quotient identity is tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}.
4
Multiply tan⁡θ\tan\theta by 12 to find the required expression's value.
-5
Multiplying the value of tan⁡θ\tan\theta (which is −512-\frac{5}{12}) by 12 yields −5-5.

Anahtar Kavram

Fundamental Trigonometric Identities
Tahmini Süre:1m 0s
Soru 4Soru →

Given that cos⁡θ=45\cos\theta = \frac{4}{5} and the terminal side of angle θ\theta lies in Quadrant IV, what is the value of tan⁡θ\tan\theta?

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Cevap: −34-\frac{3}{4}

Cevap

−34-\frac{3}{4}
The correct answer is −34-\frac{3}{4}. Since the angle θ\theta has its terminal side in Quadrant IV, its cosine is positive and its sine is negative. Using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, we find sin⁡2θ+(45)2=1\sin^2\theta + \left(\frac{4}{5}\right)^2 = 1, which simplifies to sin⁡2θ=925\sin^2\theta = \frac{9}{25}. Because sine is negative in Quadrant IV, sin⁡θ=−35\sin\theta = -\frac{3}{5}. Finally, applying the quotient identity tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}, we get tan⁡θ=−3/54/5=−34\tan\theta = \frac{-3/5}{4/5} = -\frac{3}{4}.

Adım Adım Çözüm

1
Determine the sign of sin⁡θ\sin\theta in Quadrant IV.
sin⁡θ<0\sin\theta < 0
In Quadrant IV, the x-coordinates (representing cosine) are positive, and the y-coordinates (representing sine) are negative.
2
Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 to calculate the value of sin⁡θ\sin\theta.
sin⁡θ=−35\sin\theta = -\frac{3}{5}
Substituting cos⁡θ=45\cos\theta = \frac{4}{5} gives sin⁡2θ+(45)2=1  ⟹  sin⁡2θ=1−1625=925\sin^2\theta + \left(\frac{4}{5}\right)^2 = 1 \implies \sin^2\theta = 1 - \frac{16}{25} = \frac{9}{25}. Taking the negative square root because sine is negative in Quadrant IV yields −35-\frac{3}{5}.
3
Use the quotient identity tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} to calculate tan⁡θ\tan\theta.
tan⁡θ=−34\tan\theta = -\frac{3}{4}
Dividing the value of sin⁡θ\sin\theta by cos⁡θ\cos\theta yields −3/54/5=−34\frac{-3/5}{4/5} = -\frac{3}{4}.

Anahtar Kavram

Fundamental Trigonometric Identities
Soru 5Soru →

An angle θ\theta satisfies π2<θ<π\frac{\pi}{2} < \theta < \pi and (sin⁡θ−cos⁡θ)2=179(\sin\theta - \cos\theta)^2 = \frac{17}{9}. What is the value of tan⁡θ+cot⁡θ\tan\theta + \cot\theta?

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Cevap: −94-\frac{9}{4}

Cevap

The value of the expression is −94-\frac{9}{4}
Expanding the square of the difference (sin⁡θ−cos⁡θ)2(\sin\theta - \cos\theta)^2 yields sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta. Applying the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 simplifies this expression to 1−2sin⁡θcos⁡θ1 - 2\sin\theta\cos\theta. Setting this equal to 179\frac{17}{9} and solving for the product of sine and cosine gives sin⁡θcos⁡θ=−49\sin\theta\cos\theta = -\frac{4}{9}. The target expression tan⁡θ+cot⁡θ\tan\theta + \cot\theta can be rewritten using quotient and reciprocal identities as sin⁡θcos⁡θ+cos⁡θsin⁡θ\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}, which simplifies by finding a common denominator to sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}. Substituting the value of sin⁡θcos⁡θ\sin\theta\cos\theta into this expression results in −94-\frac{9}{4}.

Adım Adım Çözüm

1
Expand the squared expression (sin⁡θ−cos⁡θ)2(\sin\theta - \cos\theta)^2 and apply the Pythagorean identity.
(sin⁡θ−cos⁡θ)2=sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ=1−2sin⁡θcos⁡θ(\sin\theta - \cos\theta)^2 = \sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 1 - 2\sin\theta\cos\theta
To express the squared binomial in terms of the product sin⁡θcos⁡θ\sin\theta\cos\theta using the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
2
Equate the expanded form to the given value and solve for sin⁡θcos⁡θ\sin\theta\cos\theta.
1−2sin⁡θcos⁡θ=179  ⟹  2sin⁡θcos⁡θ=−89  ⟹  sin⁡θcos⁡θ=−491 - 2\sin\theta\cos\theta = \frac{17}{9} \implies 2\sin\theta\cos\theta = -\frac{8}{9} \implies \sin\theta\cos\theta = -\frac{4}{9}
To find the value of the product sin⁡θcos⁡θ\sin\theta\cos\theta from the given equation.
3
Rewrite the target expression tan⁡θ+cot⁡θ\tan\theta + \cot\theta in terms of sine and cosine.
tan⁡θ+cot⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}
To simplify the target sum of ratios using quotient and reciprocal identities so that it depends only on the product sin⁡θcos⁡θ\sin\theta\cos\theta.
4
Substitute the value of sin⁡θcos⁡θ\sin\theta\cos\theta into the simplified expression.
tan⁡θ+cot⁡θ=1−4/9=−94\tan\theta + \cot\theta = \frac{1}{-4/9} = -\frac{9}{4}
To compute the final numerical value of the expression.

Anahtar Kavram

Simplifying trigonometric expressions using fundamental Pythagorean, quotient, and reciprocal identities, and solving for unknown products of trigonometric functions.

Alternatif Yöntem

Alternatively, one can find the individual values of sin⁡θ\sin\theta and cos⁡θ\cos\theta. Since (sin⁡θ−cos⁡θ)2=179(\sin\theta - \cos\theta)^2 = \frac{17}{9} and sin⁡θcos⁡θ=−49\sin\theta\cos\theta = -\frac{4}{9}, we can use the identity (sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ=19(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta = \frac{1}{9}. In Quadrant II, sin⁡θ>0\sin\theta > 0 and cos⁡θ<0\cos\theta < 0, and since −cos⁡θ>0-\cos\theta > 0, we have sin⁡θ−cos⁡θ=173\sin\theta - \cos\theta = \frac{\sqrt{17}}{3}. Solving the system of equations for sin⁡θ\sin\theta and cos⁡θ\cos\theta yields sin⁡θ=17−16\sin\theta = \frac{\sqrt{17} - 1}{6} (since it must be positive) and cos⁡θ=−17−16\cos\theta = \frac{-\sqrt{17} - 1}{6} (since it must be negative). Substituting these into tan⁡θ+cot⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} will yield the same result of −94-\frac{9}{4}, though this method involves significantly more algebraic work.
Tahmini Süre:3m 0s
Soru 6Soru →

For an angle θ\theta in the interval 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi, the expression sec⁡θ−tan⁡θ\sec\theta - \tan\theta is equal to 33. What is the value of csc⁡θ+cot⁡θ\csc\theta + \cot\theta?

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Cevap: -2

Cevap

The value of csc⁡θ+cot⁡θ\csc\theta + \cot\theta is −2-2.
By using the difference of squares on the identity sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1, we get (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1. Substituting sec⁡θ−tan⁡θ=3\sec\theta - \tan\theta = 3 gives sec⁡θ+tan⁡θ=13\sec\theta + \tan\theta = \frac{1}{3}. Solving the system of equations gives sec⁡θ=53\sec\theta = \frac{5}{3} and tan⁡θ=−43\tan\theta = -\frac{4}{3}. Since θ\theta lies in Quadrant IV, cos⁡θ=35\cos\theta = \frac{3}{5} and sin⁡θ=−45\sin\theta = -\frac{4}{5}. We then find csc⁡θ=−54\csc\theta = -\frac{5}{4} and cot⁡θ=−34\cot\theta = -\frac{3}{4}, which sum to −2-2.

Adım Adım Çözüm

1
Use the Pythagorean identity sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1, which factors into (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1.
Since sec⁡θ−tan⁡θ=3\sec\theta - \tan\theta = 3, we have 3(sec⁡θ+tan⁡θ)=1  ⟹  sec⁡θ+tan⁡θ=133(\sec\theta + \tan\theta) = 1 \implies \sec\theta + \tan\theta = \frac{1}{3}.
To establish a second linear equation in terms of sec⁡θ\sec\theta and tan⁡θ\tan\theta.
2
Add and subtract the two equations: sec⁡θ−tan⁡θ=3\sec\theta - \tan\theta = 3 and sec⁡θ+tan⁡θ=13\sec\theta + \tan\theta = \frac{1}{3}.
Adding them gives 2sec⁡θ=103  ⟹  sec⁡θ=532\sec\theta = \frac{10}{3} \implies \sec\theta = \frac{5}{3}. Subtracting the first from the second gives 2tan⁡θ=−83  ⟹  tan⁡θ=−432\tan\theta = -\frac{8}{3} \implies \tan\theta = -\frac{4}{3}.
To isolate the values of sec⁡θ\sec\theta and tan⁡θ\tan\theta.
3
Find cos⁡θ\cos\theta and sin⁡θ\sin\theta using cos⁡θ=1sec⁡θ\cos\theta = \frac{1}{\sec\theta} and sin⁡θ=tan⁡θcos⁡θ\sin\theta = \tan\theta\cos\theta.
cos⁡θ=35\cos\theta = \frac{3}{5} and sin⁡θ=−45\sin\theta = -\frac{4}{5}. Since 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi (Quadrant IV), cosine is positive and sine is negative, which matches these values.
To find the primary trigonometric values needed for the reciprocal functions.
4
Calculate csc⁡θ\csc\theta and cot⁡θ\cot\theta using reciprocal identities.
csc⁡θ=1sin⁡θ=−54\csc\theta = \frac{1}{\sin\theta} = -\frac{5}{4} and cot⁡θ=1tan⁡θ=−34\cot\theta = \frac{1}{\tan\theta} = -\frac{3}{4}.
To obtain the terms of the required sum.
5
Sum the values of csc⁡θ\csc\theta and cot⁡θ\cot\theta.
csc⁡θ+cot⁡θ=−54+(−34)=−84=−2\csc\theta + \cot\theta = -\frac{5}{4} + \left(-\frac{3}{4}\right) = -\frac{8}{4} = -2.
To obtain the final value requested by the question.

Anahtar Kavram

Pythagorean and Reciprocal Trigonometric Identities
Soru 7Soru →

For an angle θ\theta such that π<θ<3π2\pi < \theta < \frac{3\pi}{2}, the equation 1−sin⁡θcos⁡θ+cos⁡θ1−sin⁡θ=−103\frac{1 - \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = -\frac{10}{3} is satisfied. What is the value of sin⁡θ\sin\theta?

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Cevap: −45-\frac{4}{5}

Cevap

The value of sin⁡θ\sin\theta is −45-\frac{4}{5}.
The correct answer is determined by first rewriting the given equation by finding a common denominator, which simplifies the numerator to 2(1−sin⁡θ)2(1-\sin\theta) using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. The term (1−sin⁡θ)(1-\sin\theta) cancels out, resulting in 2cos⁡θ=−103\frac{2}{\cos\theta} = -\frac{10}{3}, which implies cos⁡θ=−35\cos\theta = -\frac{3}{5}. In Quadrant III, sine is negative, so using the identity sin⁡θ=−1−cos⁡2θ\sin\theta = -\sqrt{1 - \cos^2\theta} yields the correct value.

Adım Adım Çözüm

1
Find a common denominator to add the fractions on the left-hand side of the equation.
The expression becomes (1−sin⁡θ)2+cos⁡2θcos⁡θ(1−sin⁡θ)\frac{(1 - \sin\theta)^2 + \cos^2\theta}{\cos\theta(1 - \sin\theta)}.
To combine the algebraic terms into a single rational expression.
2
Expand the numerator and apply the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
The numerator simplifies to 1−2sin⁡θ+sin⁡2θ+cos⁡2θ=1−2sin⁡θ+1=2−2sin⁡θ=2(1−sin⁡θ)1 - 2\sin\theta + \sin^2\theta + \cos^2\theta = 1 - 2\sin\theta + 1 = 2 - 2\sin\theta = 2(1 - \sin\theta).
To reduce the numerator's complexity using fundamental trigonometric identities.
3
Cancel the common factor (1−sin⁡θ)(1 - \sin\theta) from the numerator and denominator, and equate the simplified term to −103-\frac{10}{3}.
The equation simplifies to 2cos⁡θ=−103\frac{2}{\cos\theta} = -\frac{10}{3}, which gives cos⁡θ=−35\cos\theta = -\frac{3}{5}.
To solve for the cosine of the angle θ\theta.
4
Use the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 to calculate sin⁡θ\sin\theta, keeping the quadrant sign rules in mind.
Since π<θ<3π2\pi < \theta < \frac{3\pi}{2} (Quadrant III), sin⁡θ\sin\theta is negative. Therefore, sin⁡θ=−1−cos⁡2θ=−1−(−35)2=−45\sin\theta = -\sqrt{1 - \cos^2\theta} = -\sqrt{1 - \left(-\frac{3}{5}\right)^2} = -\frac{4}{5}.
To find the correct value and sign of the sine ratio for the given quadrant.

Anahtar Kavram

Fundamental Trigonometric Identities
Soru 8Soru →

An angle θ\theta lies in the third quadrant, where π<θ<3π2\pi < \theta < \frac{3\pi}{2}. If sin⁡θ−cos⁡θ=−15\sin\theta - \cos\theta = -\frac{1}{5}, what is the value of the expression 125(sin⁡3θ+cos⁡3θ)125(\sin^3\theta + \cos^3\theta)?

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Cevap: -91

Cevap

The correct value of the expression is -91.
Squaring the equation sin⁡θ−cos⁡θ=−15\sin\theta - \cos\theta = -\frac{1}{5} gives 1−2sin⁡θcos⁡θ=1251 - 2\sin\theta\cos\theta = \frac{1}{25}, which simplifies to sin⁡θcos⁡θ=1225\sin\theta\cos\theta = \frac{12}{25}. We then find the square of the sum: (sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ=1+2425=4925(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta = 1 + \frac{24}{25} = \frac{49}{25}. Since the angle is in the third quadrant, both trigonometric functions are negative, so we choose the negative root sin⁡θ+cos⁡θ=−75\sin\theta + \cos\theta = -\frac{7}{5}. Factoring the sum of cubes gives sin⁡3θ+cos⁡3θ=(sin⁡θ+cos⁡θ)(sin⁡2θ−sin⁡θcos⁡θ+cos⁡2θ)=(−75)(1−1225)=−91125\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta) = (-\frac{7}{5})(1 - \frac{12}{25}) = -\frac{91}{125}. Multiplying this by 125 yields -91.

Adım Adım Çözüm

1
Square both sides of the equation sin⁡θ−cos⁡θ=−15\sin\theta - \cos\theta = -\frac{1}{5}
sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ=125  ⟹  1−2sin⁡θcos⁡θ=125  ⟹  sin⁡θcos⁡θ=1225\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = \frac{1}{25} \implies 1 - 2\sin\theta\cos\theta = \frac{1}{25} \implies \sin\theta\cos\theta = \frac{12}{25}
To solve for the product of sine and cosine using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
2
Determine the value of sin⁡θ+cos⁡θ\sin\theta + \cos\theta using the identity (sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta
(sin⁡θ+cos⁡θ)2=1+2(1225)=4925  ⟹  sin⁡θ+cos⁡θ=−75(\sin\theta + \cos\theta)^2 = 1 + 2(\frac{12}{25}) = \frac{49}{25} \implies \sin\theta + \cos\theta = -\frac{7}{5}
Because θ\theta is in the third quadrant (π<θ<3π2\pi < \theta < \frac{3\pi}{2}), both sin⁡θ\sin\theta and cos⁡θ\cos\theta must be negative, meaning their sum is also negative.
3
Use the sum of cubes factorization to evaluate sin⁡3θ+cos⁡3θ\sin^3\theta + \cos^3\theta
sin⁡3θ+cos⁡3θ=(sin⁡θ+cos⁡θ)(sin⁡2θ−sin⁡θcos⁡θ+cos⁡2θ)=(−75)(1−1225)=−91125\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta\cos\theta + \cos^2\theta) = (-\frac{7}{5})(1 - \frac{12}{25}) = -\frac{91}{125}
To express the sum of cubes in terms of the known sum and product of sine and cosine.
4
Multiply the evaluated sum of cubes by 125
125×(−91125)=−91125 \times (-\frac{91}{125}) = -91
To compute the final value of the requested expression.

Anahtar Kavram

Pythagorean trigonometric identities, quadrant sign analysis, and algebraic factorization of the sum of cubes

Alternatif Yöntem

Instead of applying algebraic identities to find the sum of cubes directly, we can solve for the individual values of sin⁡θ\sin\theta and cos⁡θ\cos\theta from the system of equations: sin⁡θ−cos⁡θ=−15\sin\theta - \cos\theta = -\frac{1}{5} and sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. Substituting sin⁡θ=cos⁡θ−15\sin\theta = \cos\theta - \frac{1}{5} into the second equation yields 2cos⁡2θ−25cos⁡θ−2425=02\cos^2\theta - \frac{2}{5}\cos\theta - \frac{24}{25} = 0, which factors as (5cos⁡θ+3)(5cos⁡θ−4)=0(5\cos\theta + 3)(5\cos\theta - 4) = 0. Since θ\theta is in Quadrant III, cos⁡θ=−35\cos\theta = -\frac{3}{5} and sin⁡θ=−45\sin\theta = -\frac{4}{5}. Evaluating 125(sin⁡3θ+cos⁡3θ)125(\sin^3\theta + \cos^3\theta) directly with these values gives 125((−45)3+(−35)3)=125(−64125−27125)=−91125((-\frac{4}{5})^3 + (-\frac{3}{5})^3) = 125(-\frac{64}{125} - \frac{27}{125}) = -91.
Tahmini Süre:2m 30s
Soru 9Soru →

For an angle θ\theta such that π2<θ<π\frac{\pi}{2} < \theta < \pi, if cos⁡θ1−sin⁡θ=−3\frac{\cos\theta}{1 - \sin\theta} = -3, what is the value of sin⁡θ−tan⁡θ\sin\theta - \tan\theta?

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Cevap: 3215\frac{32}{15}

Cevap

3215\frac{32}{15}
By multiplying the numerator and denominator of the given expression by 1+sin⁡θ1 + \sin\theta and utilizing the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, we find that sec⁡θ+tan⁡θ=−3\sec\theta + \tan\theta = -3. Since sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1, it follows that sec⁡θ−tan⁡θ=−13\sec\theta - \tan\theta = -\frac{1}{3}. Solving this system of equations yields sec⁡θ=−53\sec\theta = -\frac{5}{3} (which means cos⁡θ=−35\cos\theta = -\frac{3}{5}) and tan⁡θ=−43\tan\theta = -\frac{4}{3}. In Quadrant II, sine is positive, so sin⁡θ=tan⁡θcos⁡θ=45\sin\theta = \tan\theta \cos\theta = \frac{4}{5}. Thus, sin⁡θ−tan⁡θ=45−(−43)=12+2015=3215\sin\theta - \tan\theta = \frac{4}{5} - \left(-\frac{4}{3}\right) = \frac{12 + 20}{15} = \frac{32}{15}.

Adım Adım Çözüm

1
Multiply the numerator and denominator of the given expression by 1+sin⁡θ1 + \sin\theta to simplify it.
cos⁡θ(1+sin⁡θ)(1−sin⁡θ)(1+sin⁡θ)=−3  ⟹  cos⁡θ(1+sin⁡θ)1−sin⁡2θ=−3\frac{\cos\theta(1 + \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)} = -3 \implies \frac{\cos\theta(1 + \sin\theta)}{1 - \sin^2\theta} = -3
To set up the Pythagorean identity in the denominator.
2
Substitute the Pythagorean identity 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta into the denominator.
cos⁡θ(1+sin⁡θ)cos⁡2θ=1+sin⁡θcos⁡θ=sec⁡θ+tan⁡θ=−3\frac{\cos\theta(1 + \sin\theta)}{\cos^2\theta} = \frac{1 + \sin\theta}{\cos\theta} = \sec\theta + \tan\theta = -3
To simplify the expression into basic trigonometric functions.
3
Use the reciprocal identity relationship sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1 to find the difference of secant and tangent.
sec⁡θ−tan⁡θ=1sec⁡θ+tan⁡θ=1−3=−13\sec\theta - \tan\theta = \frac{1}{\sec\theta + \tan\theta} = \frac{1}{-3} = -\frac{1}{3}
To create a system of linear equations for secant and tangent.
4
Solve the system of equations for sec⁡θ\sec\theta and tan⁡θ\tan\theta.
sec⁡θ=−53\sec\theta = -\frac{5}{3} and tan⁡θ=−43\tan\theta = -\frac{4}{3}
Adding the two equations yields 2sec⁡θ=−103  ⟹  sec⁡θ=−532\sec\theta = -\frac{10}{3} \implies \sec\theta = -\frac{5}{3}, and subtracting them yields 2tan⁡θ=−83  ⟹  tan⁡θ=−432\tan\theta = -\frac{8}{3} \implies \tan\theta = -\frac{4}{3}.
5
Find cos⁡θ\cos\theta and sin⁡θ\sin\theta, and calculate the final value of sin⁡θ−tan⁡θ\sin\theta - \tan\theta.
cos⁡θ=−35\cos\theta = -\frac{3}{5}, sin⁡θ=45\sin\theta = \frac{4}{5}, and sin⁡θ−tan⁡θ=45−(−43)=3215\sin\theta - \tan\theta = \frac{4}{5} - \left(-\frac{4}{3}\right) = \frac{32}{15}
Since θ\theta lies in Quadrant II, sine is positive, which is verified by sin⁡θ=tan⁡θ⋅cos⁡θ=(−43)(−35)=45\sin\theta = \tan\theta \cdot \cos\theta = \left(-\frac{4}{3}\right)\left(-\frac{3}{5}\right) = \frac{4}{5}.

Anahtar Kavram

Fundamental Trigonometric Identities
Tahmini Süre:2m 0s
Soru 10Soru →

An angle θ\theta is positioned in the second quadrant. If the trigonometric expression csc⁡θ−cot⁡θ\csc\theta - \cot\theta is equal to 44, what is the value of 17cos⁡θ17\cos\theta?

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Cevap: -15

Cevap

The value of 17cos⁡θ17\cos\theta is −15-15.
The correct value is −15-15. By using the difference of squares on the identity csc⁡2θ−cot⁡2θ=1\csc^2\theta - \cot^2\theta = 1, we establish that csc⁡θ+cot⁡θ=14\csc\theta + \cot\theta = \frac{1}{4}. Solving this system alongside csc⁡θ−cot⁡θ=4\csc\theta - \cot\theta = 4 yields csc⁡θ=178\csc\theta = \frac{17}{8} and cot⁡θ=−158\cot\theta = -\frac{15}{8}. Since cos⁡θ=cot⁡θcsc⁡θ\cos\theta = \frac{\cot\theta}{\csc\theta}, we find cos⁡θ=−1517\cos\theta = -\frac{15}{17}, which means 17cos⁡θ=−1517\cos\theta = -15.

Adım Adım Çözüm

1
Use the Pythagorean identity linking cosecant and cotangent to find the sum of the terms.
csc⁡θ+cot⁡θ=14\csc\theta + \cot\theta = \frac{1}{4}
Since csc⁡2θ−cot⁡2θ=1\csc^2\theta - \cot^2\theta = 1, factoring as a difference of squares gives (csc⁡θ−cot⁡θ)(csc⁡θ+cot⁡θ)=1(\csc\theta - \cot\theta)(\csc\theta + \cot\theta) = 1. Substituting csc⁡θ−cot⁡θ=4\csc\theta - \cot\theta = 4 yields 4(csc⁡θ+cot⁡θ)=14(\csc\theta + \cot\theta) = 1, so csc⁡θ+cot⁡θ=14\csc\theta + \cot\theta = \frac{1}{4}.
2
Solve the system of equations for csc⁡θ\csc\theta and cot⁡θ\cot\theta.
csc⁡θ=178\csc\theta = \frac{17}{8} and cot⁡θ=−158\cot\theta = -\frac{15}{8}
Adding the equations csc⁡θ−cot⁡θ=4\csc\theta - \cot\theta = 4 and csc⁡θ+cot⁡θ=14\csc\theta + \cot\theta = \frac{1}{4} gives 2csc⁡θ=174  ⟹  csc⁡θ=1782\csc\theta = \frac{17}{4} \implies \csc\theta = \frac{17}{8}. Subtracting the first equation from the second gives 2cot⁡θ=−154  ⟹  cot⁡θ=−1582\cot\theta = -\frac{15}{4} \implies \cot\theta = -\frac{15}{8}.
3
Determine the value of cos⁡θ\cos\theta using reciprocal and quotient identities.
cos⁡θ=−1517\cos\theta = -\frac{15}{17}
Using the relationship cos⁡θ=cot⁡θcsc⁡θ\cos\theta = \frac{\cot\theta}{\csc\theta}, we substitute the values to find cos⁡θ=−15/817/8=−1517\cos\theta = \frac{-15/8}{17/8} = -\frac{15}{17}.
4
Calculate the final required expression value.
17cos⁡θ=−1517\cos\theta = -15
Multiplying the calculated value of cos⁡θ\cos\theta by 1717 gives 17(−1517)=−1517 \left(-\frac{15}{17}\right) = -15.

Anahtar Kavram

Using fundamental Pythagorean, reciprocal, and quotient trigonometric identities to solve systems of equations and evaluate trigonometric expressions.
Soru 11Soru →

For an angle θ\theta satisfying π2<θ<π\frac{\pi}{2} < \theta < \pi, if sec⁡θ=−135\sec \theta = -\frac{13}{5}, what is the value of 12(cot⁡θ+csc⁡θ)12(\cot \theta + \csc \theta)?

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Cevap: 8

Cevap

8
For an angle θ\theta in Quadrant II (π2<θ<π\frac{\pi}{2} < \theta < \pi), cosine is negative and sine is positive. Given sec⁡θ=−135\sec \theta = -\frac{13}{5}, the reciprocal identity gives cos⁡θ=−513\cos \theta = -\frac{5}{13}. The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 yields sin⁡θ=1−(−513)2=1213\sin \theta = \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \frac{12}{13}. Then cot⁡θ=cos⁡θsin⁡θ=−512\cot \theta = \frac{\cos \theta}{\sin \theta} = -\frac{5}{12} and csc⁡θ=1sin⁡θ=1312\csc \theta = \frac{1}{\sin \theta} = \frac{13}{12}. Adding these values gives cot⁡θ+csc⁡θ=812=23\cot \theta + \csc \theta = \frac{8}{12} = \frac{2}{3}. Multiplying by 12 yields the final value of 8.

Adım Adım Çözüm

1
Find cos⁡θ\cos \theta from sec⁡θ\sec \theta
cos⁡θ=−513\cos \theta = -\frac{5}{13}
By definition of the reciprocal trigonometric identity, cos⁡θ=1sec⁡θ\cos \theta = \frac{1}{\sec \theta}.
2
Calculate sin⁡θ\sin \theta using the Pythagorean identity
sin⁡θ=1213\sin \theta = \frac{12}{13}
In Quadrant II (π2<θ<π\frac{\pi}{2} < \theta < \pi), sine is positive. Applying sin⁡θ=1−cos⁡2θ\sin \theta = \sqrt{1 - \cos^2 \theta} gives 1−25169=1213\sqrt{1 - \frac{25}{169}} = \frac{12}{13}.
3
Find cot⁡θ\cot \theta and csc⁡θ\csc \theta
\cot \theta = -\frac{5}{12} \text{ and } \csc \theta = \frac{13}{12}
Using quotient identity cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta} and reciprocal identity csc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta}.
4
Substitute into the given expression 12(cot⁡θ+csc⁡θ)12(\cot \theta + \csc \theta) and simplify
8
12(−512+1312)=12(812)=812\left(-\frac{5}{12} + \frac{13}{12}\right) = 12\left(\frac{8}{12}\right) = 8.

Anahtar Kavram

Pythagorean, quotient, and reciprocal identities with quadrant sign analysis
Soru 12Soru →

If θ\theta is an angle in Quadrant IV such that cos⁡θ=45\cos\theta = \frac{4}{5}, what is the value of tan⁡θ+sec⁡θcsc⁡θ\frac{\tan\theta + \sec\theta}{\csc\theta}?

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Cevap: −310-\frac{3}{10}

Cevap

−310-\frac{3}{10}
Using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 in Quadrant IV yields sin⁡θ=−35\sin\theta = -\frac{3}{5}. Substituting this into quotient and reciprocal identities gives tan⁡θ=−34\tan\theta = -\frac{3}{4}, sec⁡θ=54\sec\theta = \frac{5}{4}, and csc⁡θ=−53\csc\theta = -\frac{5}{3}. Evaluating tan⁡θ+sec⁡θcsc⁡θ\frac{\tan\theta + \sec\theta}{\csc\theta} gives 1/2−5/3=−310\frac{1/2}{-5/3} = -\frac{3}{10}.

Adım Adım Çözüm

1
Determine sin⁡θ\sin\theta using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 and quadrant sign rules.
sin⁡θ=−1−(45)2=−925=−35\sin\theta = -\sqrt{1 - \left(\frac{4}{5}\right)^2} = -\sqrt{\frac{9}{25}} = -\frac{3}{5} because sine is negative in Quadrant IV.
The Pythagorean identity relates sine and cosine, and the angle's quadrant determines the sign of the trigonometric ratio.
2
Calculate the values of tan⁡θ\tan\theta, sec⁡θ\sec\theta, and csc⁡θ\csc\theta using reciprocal and quotient identities.
tan⁡θ=sin⁡θcos⁡θ=−3/54/5=−34\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-3/5}{4/5} = -\frac{3}{4}, sec⁡θ=1cos⁡θ=54\sec\theta = \frac{1}{\cos\theta} = \frac{5}{4}, and csc⁡θ=1sin⁡θ=−53\csc\theta = \frac{1}{\sin\theta} = -\frac{5}{3}.
Fundamental quotient and reciprocal identities define these functions in terms of sine and cosine.
3
Substitute the trigonometric values into the given expression tan⁡θ+sec⁡θcsc⁡θ\frac{\tan\theta + \sec\theta}{\csc\theta} and simplify.
−34+54−53=24−53=12−53=12×(−35)=−310\frac{-\frac{3}{4} + \frac{5}{4}}{-\frac{5}{3}} = \frac{\frac{2}{4}}{-\frac{5}{3}} = \frac{\frac{1}{2}}{-\frac{5}{3}} = \frac{1}{2} \times \left(-\frac{3}{5}\right) = -\frac{3}{10}.
Combining terms in the numerator and dividing by the fraction in the denominator yields the simplified value.

Anahtar Kavram

Pythagorean, quotient, and reciprocal trigonometric identities with quadrant-dependent signs
Soru 13Soru →

If π<θ<3π2\pi < \theta < \frac{3\pi}{2} and tan⁡θ=34\tan \theta = \frac{3}{4}, what is the value of the expression sin⁡2θ1+cos⁡θ\frac{\sin^2 \theta}{1 + \cos \theta}?

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Cevap: 95\frac{9}{5}

Cevap

95\frac{9}{5}
The expression sin⁡2θ1+cos⁡θ\frac{\sin^2 \theta}{1 + \cos \theta} can be simplified using the Pythagorean identity sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta. Factoring the numerator gives (1−cos⁡θ)(1+cos⁡θ)(1 - \cos \theta)(1 + \cos \theta), which cancels with the denominator to leave 1−cos⁡θ1 - \cos \theta. Given tan⁡θ=34\tan \theta = \frac{3}{4} in Quadrant III, the reference right triangle has sides 3 and 4 with hypotenuse 5. Since cosine is negative in the third quadrant, cos⁡θ=−45\cos \theta = -\frac{4}{5}. Evaluating 1−(−45)1 - \left(-\frac{4}{5}\right) yields 95\frac{9}{5}.

Adım Adım Çözüm

1
Simplify the algebraic trigonometric expression using standard fundamental identities.
sin⁡2θ1+cos⁡θ=1−cos⁡2θ1+cos⁡θ=(1−cos⁡θ)(1+cos⁡θ)1+cos⁡θ=1−cos⁡θ\frac{\sin^2 \theta}{1 + \cos \theta} = \frac{1 - \cos^2 \theta}{1 + \cos \theta} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 + \cos \theta} = 1 - \cos \theta
Applying the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 allows factoring and canceling terms.
2
Determine the value and sign of cos⁡θ\cos \theta based on the given tangent ratio and quadrant constraint.
cos⁡θ=−45\cos \theta = -\frac{4}{5}
Since tan⁡θ=34=oppositeadjacent\tan \theta = \frac{3}{4} = \frac{\text{opposite}}{\text{adjacent}}, the hypotenuse is 55. In Quadrant III (π<θ<3π2\pi < \theta < \frac{3\pi}{2}), cosine is negative.
3
Substitute the value of cos⁡θ\cos \theta into the simplified expression.
1−(−45)=1+45=951 - \left(-\frac{4}{5}\right) = 1 + \frac{4}{5} = \frac{9}{5}
Subtracting a negative value results in addition.

Anahtar Kavram

Fundamental Pythagorean Identities and Quadrant Signs of Trigonometric Functions
Tahmini Süre:1m 15s
Soru 14Soru →

If sin⁡θ−cos⁡θ=0.6\sin \theta - \cos \theta = 0.6, what is the value of sin⁡θcos⁡θ\sin \theta \cos \theta?

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Cevap: 0.32

Cevap

The value of sin⁡θcos⁡θ\sin \theta \cos \theta is 0.32.
Squaring both sides of sin⁡θ−cos⁡cos⁡θ=0.6\sin \theta - \cos \cos \theta = 0.6 yields sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ=0.36\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = 0.36. Substituting the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 gives 1−2sin⁡θcos⁡θ=0.361 - 2\sin \theta \cos \theta = 0.36, which rearranges to 2sin⁡θcos⁡θ=0.642\sin \theta \cos \theta = 0.64. Dividing by 2 yields sin⁡θcos⁡θ=0.32\sin \theta \cos \theta = 0.32.

Adım Adım Çözüm

1
Square both sides of the given equation
(sin⁡θ−cos⁡θ)2=0.36(\sin \theta - \cos \theta)^2 = 0.36
Squaring allows us to introduce the product sin⁡θcos⁡θ\sin \theta \cos \theta alongside sin⁡2θ\sin^2 \theta and cos⁡2θ\cos^2 \theta.
2
Expand the binomial on the left side
sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ=0.36\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = 0.36
Use the algebraic expansion identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.
3
Substitute the fundamental Pythagorean trigonometric identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1
1−2sin⁡θcos⁡θ=0.361 - 2\sin \theta \cos \theta = 0.36
sin⁡2θ+cos⁡2θ\sin^2 \theta + \cos^2 \theta always equals 1 for any angle θ\theta.
4
Isolate the term containing sin⁡θcos⁡θ\sin \theta \cos \theta
2sin⁡θcos⁡θ=0.642\sin \theta \cos \theta = 0.64
Subtract 0.36 from 1 to find the value of 2sin⁡θcos⁡θ2\sin \theta \cos \theta.
5
Divide by 2 to solve for sin⁡θcos⁡θ\sin \theta \cos \theta
sin⁡θcos⁡θ=0.32\sin \theta \cos \theta = 0.32
Simplifies 0.64/20.64 / 2 to obtain the final required numerical value.

Anahtar Kavram

Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1
Soru 15Soru →

For an angle θ\theta in the third quadrant satisfying π<θ<3π2\pi < \theta < \frac{3\pi}{2}, the tangent value is tan⁡θ=43\tan \theta = \frac{4}{3}. What is the exact value of the expression sin⁡4θ−cos⁡4θ\sin^4 \theta - \cos^4 \theta?

Cevabı ve açıklamayı göster

Cevap: 0.28

Cevap

The exact numerical value of the expression is 0.28 (or 7/25).
By factoring sin⁡4θ−cos⁡4θ\sin^4 \theta - \cos^4 \theta as (sin⁡2θ−cos⁡2θ)(sin⁡2θ+cos⁡2θ)(\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta), we can apply the fundamental Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1. The expression simplifies cleanly to sin⁡2θ−cos⁡2θ\sin^2 \theta - \cos^2 \theta. Given tan⁡θ=43\tan \theta = \frac{4}{3} in Quadrant III, the reference triangle has opposite side 4, adjacent side 3, and hypotenuse 5. Thus, sin⁡θ=−45\sin \theta = -\frac{4}{5} and cos⁡θ=−35\cos \theta = -\frac{3}{5}. Substituting these values yields (−45)2−(−35)2=1625−925=725=0.28\left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} = 0.28.

Adım Adım Çözüm

1
Factor the fourth-degree trigonometric expression using difference of squares.
sin⁡4θ−cos⁡4θ=(sin⁡2θ−cos⁡2θ)(sin⁡2θ+cos⁡2θ)\sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)
The difference of two squares a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b) applies directly to a=sin⁡2θa = \sin^2 \theta and b=cos⁡2θb = \cos^2 \theta.
2
Simplify using the fundamental Pythagorean trigonometric identity.
sin⁡4θ−cos⁡4θ=sin⁡2θ−cos⁡2θ\sin^4 \theta - \cos^4 \theta = \sin^2 \theta - \cos^2 \theta
By the Pythagorean identity, sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 for any angle θ\theta.
3
Determine sin⁡θ\sin \theta and cos⁡θ\cos \theta from the given quadrant and tangent value.
sin⁡θ=−45\sin \theta = -\frac{4}{5} and cos⁡θ=−35\cos \theta = -\frac{3}{5}
In Quadrant III (π<θ<3π2\pi < \theta < \frac{3\pi}{2}), both sine and cosine are negative. A standard 3-4-5 right triangle yields sin⁡θ=−45\sin \theta = -\frac{4}{5} and cos⁡θ=−35\cos \theta = -\frac{3}{5}.
4
Substitute the values into the simplified expression and compute the result.
(−45)2−(−35)2=1625−925=725=0.28\left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} = 0.28
Squaring each trigonometric ratio yields positive values, giving a final simplified decimal value of 0.28.

Anahtar Kavram

Pythagorean Identity and Difference of Squares
Tahmini Süre:1m 30s
Soru 16Soru →

For an angle θ\theta satisfying π<θ<3π2\pi < \theta < \frac{3\pi}{2}, if cos⁡θ=−1213\cos \theta = -\frac{12}{13}, what is the value of sec⁡θ−cos⁡θtan⁡θ\frac{\sec \theta - \cos \theta}{\tan \theta}?

Cevabı ve açıklamayı göster

Cevap: −513-\frac{5}{13}

Cevap

The expression simplifies to sin⁡θ\sin \theta, which equals −513-\frac{5}{13}.
Using identity substitutions sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}, 1−cos⁡2θ=sin⁡2θ1 - \cos^2 \theta = \sin^2 \theta, and tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}, the expression simplifies directly to sin⁡θ\sin \theta. Since cos⁡θ=−1213\cos \theta = -\frac{12}{13} in Quadrant III where sine is negative, sin⁡θ=−1−(−1213)2=−513\sin \theta = -\sqrt{1 - \left(-\frac{12}{13}\right)^2} = -\frac{5}{13}.

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1
Simplify the given trigonometric expression using fundamental identities.
sec⁡θ−cos⁡θtan⁡θ=1cos⁡θ−cos⁡θsin⁡θcos⁡θ=1−cos⁡2θcos⁡θsin⁡θcos⁡θ=sin⁡2θsin⁡θ=sin⁡θ\frac{\sec \theta - \cos \theta}{\tan \theta} = \frac{\frac{1}{\cos \theta} - \cos \theta}{\frac{\sin \theta}{\cos \theta}} = \frac{\frac{1 - \cos^2 \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = \frac{\sin^2 \theta}{\sin \theta} = \sin \theta
Applying the reciprocal identity sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}, the quotient identity tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}, and the Pythagorean identity 1−cos⁡2θ=sin⁡2θ1 - \cos^2 \theta = \sin^2 \theta simplifies the expression directly to sin⁡θ\sin \theta.
2
Calculate the magnitude of sin⁡θ\sin \theta using the Pythagorean identity.
∣sin⁡θ∣=1−cos⁡2θ=1−(−1213)2=1−144169=25169=513|\sin \theta| = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(-\frac{12}{13}\right)^2} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}
The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 allows finding the absolute value of sin⁡θ\sin \theta.
3
Determine the correct sign for sin⁡θ\sin \theta based on the given quadrant.
sin⁡θ=−513\sin \theta = -\frac{5}{13}
Since π<θ<3π2\pi < \theta < \frac{3\pi}{2}, the angle lies in Quadrant III, where the sine function is negative.

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Simplifying expressions using fundamental Pythagorean, reciprocal, and quotient identities while applying quadrant sign rules.
Soru 17Soru →

If sin⁡θ+cos⁡θ=1.5\sin \theta + \cos \theta = \sqrt{1.5}, what is the value of tan⁡θ+cot⁡θ\tan \theta + \cot \theta?

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Cevap: 4

Cevap

4
Squaring both sides of sin⁡θ+cos⁡θ=1.5\sin \theta + \cos \theta = \sqrt{1.5} yields sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=1.5\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1.5. Using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, we get 1+2sin⁡θcos⁡θ=1.51 + 2\sin \theta \cos \theta = 1.5, which simplifies to sin⁡θcos⁡θ=0.25\sin \theta \cos \theta = 0.25. Rewriting tan⁡θ+cot⁡θ\tan \theta + \cot \theta using quotient identities gives sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}. Substituting sin⁡θcos⁡θ=0.25\sin \theta \cos \theta = 0.25 yields 10.25=4\frac{1}{0.25} = 4.

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1
Square both sides of the given equation
sin⁡2θ+2sin⁡θcos⁡θ+cos⁡2θ=1.5\sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1.5
Squaring both sides allows the expansion of the binomial (sin⁡θ+cos⁡θ)2(\sin \theta + \cos \theta)^2 to reveal the product term sin⁡θcos⁡θ\sin \theta \cos \theta.
2
Apply the Pythagorean identity to solve for sin⁡θcos⁡θ\sin \theta \cos \theta
sin⁡θcos⁡θ=0.25\sin \theta \cos \theta = 0.25
Since sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, substituting 11 into 1+2sin⁡θcos⁡θ=1.51 + 2\sin \theta \cos \theta = 1.5 gives 2sin⁡θcos⁡θ=0.52\sin \theta \cos \theta = 0.5, so sin⁡θcos⁡θ=0.25\sin \theta \cos \theta = 0.25.
3
Rewrite tan⁡θ+cot⁡θ\tan \theta + \cot \theta using quotient identities
tan⁡θ+cot⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}
Using tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} and cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}, finding a common denominator yields sin⁡2θ+cos⁡2θsin⁡θcos⁡θ\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}. Applying sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 simplifies the numerator to 11.
4
Substitute the numerical value of sin⁡θcos⁡θ\sin \theta \cos \theta to find the answer
tan⁡θ+cot⁡θ=10.25=4\tan \theta + \cot \theta = \frac{1}{0.25} = 4
Dividing 11 by 0.250.25 evaluates to the exact integer 44.

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Fundamental Trigonometric Identities (Pythagorean and Quotient Identities)
Tahmini Süre:1m 30s
Soru 18Soru →

Which of the following expressions is equivalent to tan⁡θ+cot⁡θcsc⁡θ\frac{\tan \theta + \cot \theta}{\csc \theta} for all values of θ\theta where the expression is defined?

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Cevap: sec⁡θ\sec \theta

Cevap

sec⁡θ\sec \theta
Substituting tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} and cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta} gives sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} in the numerator. Dividing by csc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta} cancels the sin⁡θ\sin \theta term, leaving 1cos⁡θ\frac{1}{\cos \theta}, which is equal to sec⁡θ\sec \theta.

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1
Express tangent and cotangent using sine and cosine
tan⁡θ+cot⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}
Quotient identities allow rewriting all functions in terms of sine and cosine.
2
Combine the fractions in the numerator over a common denominator
\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta}
Finding a common denominator of sin⁡θcos⁡θ\sin \theta \cos \theta allows application of the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.
3
Divide by the denominator csc⁡θ\csc \theta
\frac{\frac{1}{\sin \theta \cos \theta}}{\frac{1}{\sin \theta}} = \frac{1}{\sin \theta \cos \theta} \cdot \frac{\sin \theta}{1} = \frac{1}{\cos \theta} = \sec \theta
The cosecant function is the reciprocal of sine, so dividing by csc⁡θ\csc \theta cancels out the sin⁡θ\sin \theta factor in the denominator.

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Fundamental Trigonometric Identities
Soru 19Soru →

If tan⁡θ=2\tan \theta = 2, what is the value of 3sin⁡θ+4cos⁡θ5sin⁡θ−2cos⁡θ\frac{3\sin \theta + 4\cos \theta}{5\sin \theta - 2\cos \theta}?

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Cevap: 1.25

Cevap

The value of the expression is 1.25.
Using the trigonometric quotient identity tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}, dividing both the numerator and denominator of 3sin⁡θ+4cos⁡θ5sin⁡θ−2cos⁡θ\frac{3\sin \theta + 4\cos \theta}{5\sin \theta - 2\cos \theta} by cos⁡θ\cos \theta transforms the fraction into 3tan⁡θ+45tan⁡θ−2\frac{3\tan \theta + 4}{5\tan \theta - 2}. Substituting tan⁡θ=2\tan \theta = 2 gives 3(2)+45(2)−2=108=1.25\frac{3(2) + 4}{5(2) - 2} = \frac{10}{8} = 1.25. Alternatively, constructing a right triangle with opposite side 2 and adjacent side 1 gives a hypotenuse of 5\sqrt{5}, yielding sin⁡θ=25\sin \theta = \frac{2}{\sqrt{5}} and cos⁡θ=15\cos \theta = \frac{1}{\sqrt{5}}, which produces the exact same ratio of 10/58/5=1.25\frac{10/\sqrt{5}}{8/\sqrt{5}} = 1.25.

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1
Divide every term in both the numerator and the denominator by cos⁡θ\cos \theta.
The expression becomes 3(sin⁡θcos⁡θ)+4(cos⁡θcos⁡θ)5(sin⁡θcos⁡θ)−2(cos⁡θcos⁡θ)\frac{3\left(\frac{\sin \theta}{\cos \theta}\right) + 4\left(\frac{\cos \theta}{\cos \theta}\right)}{5\left(\frac{\sin \theta}{\cos \theta}\right) - 2\left(\frac{\cos \theta}{\cos \theta}\right)}.
Dividing by cos⁡θ\cos \theta allows us to convert sine-and-cosine terms into tangent terms using the quotient identity.
2
Apply the quotient identity tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}.
The expression simplifies to 3tan⁡θ+45tan⁡θ−2\frac{3\tan \theta + 4}{5\tan \theta - 2}.
This reduces the trigonometric expression to an algebraic expression containing only tan⁡θ\tan \theta.
3
Substitute tan⁡θ=2\tan \theta = 2 into the simplified expression and compute the result.
\frac{3(2) + 4}{5(2) - 2} = \frac{6 + 4}{10 - 2} = \frac{10}{8} = 1.25.
Performing basic arithmetic yields the exact numeric answer.

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Quotient Identity of Tangent