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Inverse Trigonometric Functions

10 soru

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Soru 1Soru →

What is the exact value of cos⁡(arcsin⁡(35))\cos\left(\arcsin\left(\frac{3}{5}\right)\right) expressed as a decimal?

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Cevap: 0.8

Cevap

The exact value of the expression is 0.8.
Evaluating cos⁡(arcsin⁡(35))\cos\left(\arcsin\left(\frac{3}{5}\right)\right) requires finding the cosine of an angle θ\theta whose sine is 35\frac{3}{5}. In a right triangle with an opposite side of 3 and a hypotenuse of 5, the adjacent side is 52−32=4\sqrt{5^2 - 3^2} = 4. The cosine of θ\theta is the ratio of the adjacent side to the hypotenuse, which gives 45=0.8\frac{4}{5} = 0.8.

Adım Adım Çözüm

1
Interpret the inverse sine function as an angle.
Let θ=arcsin⁡(35)\theta = \arcsin\left(\frac{3}{5}\right), meaning sin⁡(θ)=35\sin(\theta) = \frac{3}{5} for 0<θ<π20 < \theta < \frac{\pi}{2}.
The inverse sine function returns an angle whose sine is the given value within the principal interval [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
2
Find the adjacent side of the right triangle associated with angle θ\theta.
adjacent=52−32=16=4\text{adjacent} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4.
By the Pythagorean theorem, a2+b2=c2a^2 + b^2 = c^2, so the adjacent side length is c2−b2\sqrt{c^2 - b^2}.
3
Calculate the cosine of angle θ\theta.
cos⁡(θ)=adjacenthypotenuse=45=0.8\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5} = 0.8.
Cosine is defined as the ratio of the adjacent side to the hypotenuse in a right triangle.

Anahtar Kavram

Composition of Trigonometric and Inverse Trigonometric Functions
Tahmini Süre:45s
Soru 2Soru →

If θ=arcsin⁡(35)\theta = \arcsin\left(\frac{3}{5}\right), where 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2}, what is the value of tan⁡(θ)\tan(\theta)?

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Cevap: 34\frac{3}{4}

Cevap

34\frac{3}{4}
Since θ=arcsin⁡(35)\theta = \arcsin\left(\frac{3}{5}\right), sin⁡(θ)=35\sin(\theta) = \frac{3}{5}. In a right-angled triangle with acute angle θ\theta, the opposite side is 3 and the hypotenuse is 5. By the Pythagorean theorem, the adjacent side is 52−32=4\sqrt{5^2 - 3^2} = 4. The tangent of θ\theta is defined as the ratio of the opposite side to the adjacent side, which gives 34\frac{3}{4}.

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1
Interpret the inverse trigonometric equation
sin⁡(θ)=35\sin(\theta) = \frac{3}{5}
By definition of inverse sine, θ=arcsin⁡(35)\theta = \arcsin\left(\frac{3}{5}\right) means sin⁡(θ)=35\sin(\theta) = \frac{3}{5} for 0≤θ≤π20 \leq \theta \leq \frac{\pi}{2}.
2
Determine the side lengths of the reference right triangle
\text{opposite} = 3, \quad \text{hypotenuse} = 5, \quad \text{adjacent} = \sqrt{5^2 - 3^2} = 4
Using the ratio definition sin⁡(θ)=oppositehypotenuse\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} and the Pythagorean theorem a2+b2=c2a^2 + b^2 = c^2 to solve for the adjacent side.
3
Evaluate tan⁡(θ)\tan(\theta)
tan⁡(θ)=oppositeadjacent=34\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}
The tangent function is defined as the ratio of the length of the opposite side to the length of the adjacent side.

Anahtar Kavram

Evaluating composite trigonometric expressions involving inverse functions using right triangle geometry.
Soru 3Soru →

What is the exact value of cos⁡(2arctan⁡(−3))\cos\left(2\arctan(-3)\right) expressed as a decimal?

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Cevap: -0.8

Cevap

The exact decimal value of cos⁡(2arctan⁡(−3))\cos\left(2\arctan(-3)\right) is −0.8-0.8.
Letting θ=arctan⁡(−3)\theta = \arctan(-3) gives tan⁡(θ)=−3\tan(\theta) = -3. Utilizing the double-angle formula cos⁡(2θ)=1−tan⁡2(θ)1+tan⁡2(θ)\cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}, we substitute tan⁡(θ)=−3\tan(\theta) = -3 to obtain 1−91+9=−810=−0.8\frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8.

Adım Adım Çözüm

1
Define an angle variable for the inverse trigonometric expression.
Let θ=arctan⁡(−3)\theta = \arctan(-3), which means tan⁡(θ)=−3\tan(\theta) = -3 in Quadrant IV where −π2<θ<0-\frac{\pi}{2} < \theta < 0.
Applying the standard definition and principal range of the arctangent function.
2
Apply the double-angle identity for cosine expressed in terms of tangent.
cos⁡(2θ)=1−tan⁡2(θ)1+tan⁡2(θ)\cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}
This identity directly connects cos⁡(2θ)\cos(2\theta) to the known value of tan⁡(θ)\tan(\theta) without needing radicals.
3
Substitute tan⁡(θ)=−3\tan(\theta) = -3 into the identity and evaluate.
cos⁡(2θ)=1−91+9=−810=−0.8\cos(2\theta) = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8
Simplifying the numerical expression produces the exact decimal value.

Anahtar Kavram

Inverse Trigonometric Functions and Double-Angle Identities
Tahmini Süre:1m 30s
Soru 4Soru →

If α=arccos⁡(−513)\alpha = \arccos\left(-\frac{5}{13}\right), what is the exact value of tan⁡(α−π4)\tan\left(\alpha - \frac{\pi}{4}\right)?

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Cevap: 177\frac{17}{7}

Cevap

The exact value of tan⁡(α−π4)\tan\left(\alpha - \frac{\pi}{4}\right) is 177\frac{17}{7}.
Evaluating α=arccos⁡(−513)\alpha = \arccos\left(-\frac{5}{13}\right) places α\alpha in Quadrant II (π2<α<π\frac{\pi}{2} < \alpha < \pi). In this quadrant, cosine is negative (−513-\frac{5}{13}) and sine is positive (1213\frac{12}{13}), giving tan⁡α=−125\tan\alpha = -\frac{12}{5}. Substituting tan⁡α=−125\tan\alpha = -\frac{12}{5} and tan⁡(π4)=1\tan\left(\frac{\pi}{4}\right) = 1 into the identity tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} yields −125−11−125=−175−75=177\frac{-\frac{12}{5} - 1}{1 - \frac{12}{5}} = \frac{-\frac{17}{5}}{-\frac{7}{5}} = \frac{17}{7}.

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1
Determine the quadrant and reference values for α=arccos⁡(−513)\alpha = \arccos\left(-\frac{5}{13}\right)
Since the principal range of arccos⁡(x)\arccos(x) is [0,π][0, \pi] and −513<0-\frac{5}{13} < 0, α\alpha lies in Quadrant II where cos⁡α=−513\cos\alpha = -\frac{5}{13} and sin⁡α>0\sin\alpha > 0.
Inverse cosine returns angles in Quadrant II for negative arguments.
2
Calculate sin⁡α\sin\alpha and tan⁡α\tan\alpha
sin⁡α=1−(−513)2=144169=1213\sin\alpha = \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \sqrt{\frac{144}{169}} = \frac{12}{13}, so tan⁡α=sin⁡αcos⁡α=12/13−5/13=−125\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{12/13}{-5/13} = -\frac{12}{5}.
Using the Pythagorean identity and definition of tangent.
3
Apply the tangent angle subtraction identity
tan⁡(α−π4)=tan⁡α−tan⁡(π4)1+tan⁡αtan⁡(π4)=−125−11+(−125)(1)=−175−75=177\tan\left(\alpha - \frac{\pi}{4}\right) = \frac{\tan\alpha - \tan\left(\frac{\pi}{4}\right)}{1 + \tan\alpha \tan\left(\frac{\pi}{4}\right)} = \frac{-\frac{12}{5} - 1}{1 + \left(-\frac{12}{5}\right)(1)} = \frac{-\frac{17}{5}}{-\frac{7}{5}} = \frac{17}{7}.
Evaluating tan⁡(π4)=1\tan\left(\frac{\pi}{4}\right) = 1 and simplifying the complex fraction.

Anahtar Kavram

Inverse Trigonometric Functions and Compound Angle Identities
Tahmini Süre:2m 0s
Soru 5Soru →

For any real number xx such that 0<x<30 < x < 3, which of the following expressions is equivalent to sin⁡(arccos⁡(x3))\sin\left(\arccos\left(\frac{x}{3}\right)\right)?

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Cevap: 9−x23\frac{\sqrt{9 - x^2}}{3}

Cevap

9−x23\frac{\sqrt{9 - x^2}}{3}
By letting θ=arccos⁡(x3)\theta = \arccos\left(\frac{x}{3}\right), we construct a right triangle with an adjacent side of length xx and a hypotenuse of length 33. Applying the Pythagorean theorem yields an opposite side length of 32−x2=9−x2\sqrt{3^2 - x^2} = \sqrt{9 - x^2}. The sine of θ\theta is the ratio of the opposite side to the hypotenuse, which evaluates to 9−x23\frac{\sqrt{9 - x^2}}{3}.

Adım Adım Çözüm

1
Define an angle variable for the inverse trigonometric expression
Let θ=arccos⁡(x3)\theta = \arccos\left(\frac{x}{3}\right), so cos⁡(θ)=x3\cos(\theta) = \frac{x}{3} for 0<θ<π20 < \theta < \frac{\pi}{2}.
Setting the inverse trigonometric function to an angle allows setting up a right triangle relationship.
2
Set up side lengths of a right triangle using the definition of cosine
Adjacent side = xx, Hypotenuse = 33.
By definition, cos⁡(θ)=adjacenthypotenuse\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}.
3
Calculate the opposite side using the Pythagorean theorem
\text{opposite} = \sqrt{3^2 - x^2} = \sqrt{9 - x^2}.
In any right triangle, opposite2+adjacent2=hypotenuse2\text{opposite}^2 + \text{adjacent}^2 = \text{hypotenuse}^2.
4
Determine the sine of the angle
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{9 - x^2}}{3}.
By definition, sin⁡(θ)=oppositehypotenuse\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}.

Anahtar Kavram

Evaluating algebraic compositions of trigonometric and inverse trigonometric functions using right triangle geometry.
Tahmini Süre:1m 15s
Soru 6Soru →

What is the value, in degrees, of the expression arcsin⁡(32)+arccos⁡(−12)\arcsin\left(\frac{\sqrt{3}}{2}\right) + \arccos\left(-\frac{1}{2}\right)?

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Cevap: 180

Cevap

The value of the expression is 180 degrees.
To evaluate arcsin⁡(32)+arccos⁡(−12)\arcsin\left(\frac{\sqrt{3}}{2}\right) + \arccos\left(-\frac{1}{2}\right), find the principal value for each term. The principal range for arcsin⁡\arcsin is [−90∘,90∘][-90^\circ, 90^\circ], so arcsin⁡(32)=60∘\arcsin\left(\frac{\sqrt{3}}{2}\right) = 60^\circ. The principal range for arccos⁡\arccos is [0∘,180∘][0^\circ, 180^\circ], so for a negative argument, the result lies in Quadrant II, giving arccos⁡(−12)=120∘\arccos\left(-\frac{1}{2}\right) = 120^\circ. Adding these values together gives 60∘+120∘=180∘60^\circ + 120^\circ = 180^\circ.

Adım Adım Çözüm

1
Find the principal angle for arcsin⁡(32)\arcsin\left(\frac{\sqrt{3}}{2}\right) in degrees.
arcsin⁡(32)=60∘\arcsin\left(\frac{\sqrt{3}}{2}\right) = 60^\circ
The inverse sine function yields outputs restricted to the range [−90∘,90∘][-90^\circ, 90^\circ]. The angle in Quadrant I whose sine is 32\frac{\sqrt{3}}{2} is 60∘60^\circ.
2
Find the principal angle for arccos⁡(−12)\arccos\left(-\frac{1}{2}\right) in degrees.
arccos⁡(−12)=120∘\arccos\left(-\frac{1}{2}\right) = 120^\circ
The inverse cosine function yields outputs restricted to the range [0∘,180∘][0^\circ, 180^\circ]. For a negative input, the output must be in Quadrant II. The angle whose cosine is −12-\frac{1}{2} is 120∘120^\circ.
3
Sum the two evaluated angle measures.
60∘+120∘=180∘60^\circ + 120^\circ = 180^\circ
Perform standard addition on the two principal values.

Anahtar Kavram

Evaluating inverse trigonometric functions within their standard principal value ranges.
Soru 7Soru →

What is the exact value of csc⁡(arctan⁡(−724))\csc\left(\arctan\left(-\frac{7}{24}\right)\right)?

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Cevap: −257-\frac{25}{7}

Cevap

−257-\frac{25}{7}
Let θ=arctan⁡(−724)\theta = \arctan\left(-\frac{7}{24}\right). By definition of the inverse tangent function's principal range (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), a negative input produces an angle in Quadrant IV. In Quadrant IV, the opposite side is −7-7, the adjacent side is 2424, and the hypotenuse is 242+(−7)2=25\sqrt{24^2 + (-7)^2} = 25. Cosecant is the ratio of hypotenuse to opposite, giving 25−7=−257\frac{25}{-7} = -\frac{25}{7}.

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1
Determine the quadrant for θ=arctan⁡(−724)\theta = \arctan\left(-\frac{7}{24}\right)
θ\theta lies in Quadrant IV because the principal range of arctan⁡(x)\arctan(x) is (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) and the input is negative.
Inverse tangent maps negative real numbers to angles in the interval (−π2,0)\left(-\frac{\pi}{2}, 0\right).
2
Set up a right triangle ratio for tan⁡(θ)=−724\tan(\theta) = -\frac{7}{24}
Opposite side = −7-7, adjacent side = 2424, hypotenuse = 242+(−7)2=25\sqrt{24^2 + (-7)^2} = 25.
Tangent is the ratio of opposite to adjacent sides, and the hypotenuse is determined using the Pythagorean theorem.
3
Calculate csc⁡(θ)\csc(\theta)
csc⁡(θ)=hypotenuseopposite=25−7=−257\csc(\theta) = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{25}{-7} = -\frac{25}{7}.
Cosecant is defined as the reciprocal of sine, which equals hypotenuse divided by opposite side.

Anahtar Kavram

Evaluating trigonometric compositions involving inverse trigonometric functions using right triangle geometry and principal angle ranges.
Soru 8Soru →

If θ=arcsin⁡(−35)\theta = \arcsin\left(-\frac{3}{5}\right), what is the exact decimal value of cos⁡(2θ)\cos(2\theta)?

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Cevap: 0.28

Cevap

The exact decimal value of cos⁡(2θ)\cos(2\theta) is 0.280.28.
Given θ=arcsin⁡(−35)\theta = \arcsin\left(-\frac{3}{5}\right), the sine of θ\theta is sin⁡(θ)=−35\sin(\theta) = -\frac{3}{5}. Using the double-angle formula for cosine, cos⁡(2θ)=1−2sin⁡2(θ)\cos(2\theta) = 1 - 2\sin^2(\theta), we substitute sin⁡(θ)\sin(\theta) to get cos⁡(2θ)=1−2(−35)2=1−1825=725=0.28\cos(2\theta) = 1 - 2\left(-\frac{3}{5}\right)^2 = 1 - \frac{18}{25} = \frac{7}{25} = 0.28.

Adım Adım Çözüm

1
Identify the value of sin⁡(θ)\sin(\theta) from the inverse trigonometric expression
sin⁡(θ)=−35\sin(\theta) = -\frac{3}{5}
By definition of the inverse sine function, if θ=arcsin⁡(−35)\theta = \arcsin\left(-\frac{3}{5}\right), then sin⁡(θ)=−35\sin(\theta) = -\frac{3}{5} where −π2≤θ≤π2-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}.
2
Select the double-angle identity for cosine that uses sine
cos⁡(2θ)=1−2sin⁡2(θ)\cos(2\theta) = 1 - 2\sin^2(\theta)
This form of the double-angle identity allows direct calculation without needing to calculate cos⁡(θ)\cos(\theta) first.
3
Substitute sin⁡(θ)\sin(\theta) and evaluate the expression
cos⁡(2θ)=0.28\cos(2\theta) = 0.28
Substituting sin⁡(θ)=−35\sin(\theta) = -\frac{3}{5} gives 1−2(−35)2=1−2(925)=1−1825=725=0.281 - 2\left(-\frac{3}{5}\right)^2 = 1 - 2\left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{7}{25} = 0.28.

Anahtar Kavram

Evaluating Trigonometric Functions of Inverse Trigonometric Expressions using Double-Angle Identities

Alternatif Yöntem

Alternatively, place θ\theta in Quadrant IV (since −π2≤θ<0-\frac{\pi}{2} \le \theta < 0 for a negative inverse sine input). The adjacent side is 52−(−3)2=4\sqrt{5^2 - (-3)^2} = 4, so cos⁡(θ)=45\cos(\theta) = \frac{4}{5}. Then apply the alternative double-angle identity cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)=(45)2−(−35)2=1625−925=725=0.28\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \left(\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} = 0.28.
Tahmini Süre:1m 15s
Soru 9Soru →

If θ=arctan⁡(−43)\theta = \arctan\left(-\frac{4}{3}\right), what is the exact value of sin⁡(2θ)\sin(2\theta)?

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Cevap: −2425-\frac{24}{25}

Cevap

The exact value of sin⁡(2θ)\sin(2\theta) is −2425-\frac{24}{25}.
The angle θ=arctan⁡(−43)\theta = \arctan\left(-\frac{4}{3}\right) lies in Quadrant IV (−π2<θ<0-\frac{\pi}{2} < \theta < 0) because the range of arctan⁡(x)\arctan(x) is (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right). Using a right triangle in Quadrant IV with opposite side −4-4 and adjacent side 33, the hypotenuse is 55. Hence, sin⁡(θ)=−45\sin(\theta) = -\frac{4}{5} and cos⁡(θ)=35\cos(\theta) = \frac{3}{5}. Using the double-angle identity sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta), we get 2(−45)(35)=−24252\left(-\frac{4}{5}\right)\left(\frac{3}{5}\right) = -\frac{24}{25}.

Adım Adım Çözüm

1
Determine the quadrant and trigonometric ratios for θ\theta
Since θ=arctan⁡(−43)\theta = \arctan\left(-\frac{4}{3}\right), the angle θ\theta is in Quadrant IV where −π2<θ<0-\frac{\pi}{2} < \theta < 0. In this quadrant, the opposite side is −4-4, the adjacent side is 33, and the hypotenuse is 32+(−4)2=5\sqrt{3^2 + (-4)^2} = 5. Therefore, sin⁡(θ)=−45\sin(\theta) = -\frac{4}{5} and cos⁡(θ)=35\cos(\theta) = \frac{3}{5}.
The range of the principal arctangent function is (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), so a negative input places the angle in Quadrant IV.
2
Apply the double-angle formula for sine
sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)=2(−45)(35)=−2425\sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2\left(-\frac{4}{5}\right)\left(\frac{3}{5}\right) = -\frac{24}{25}.
The double-angle identity for sine expresses sin⁡(2θ)\sin(2\theta) in terms of sin⁡(θ)\sin(\theta) and cos⁡(θ)\cos(\theta).

Anahtar Kavram

Evaluating trigonometric functions of double angles involving inverse trigonometric functions
Tahmini Süre:1m 15s
Soru 10Soru →

What is the exact decimal value of tan⁡(arcsin⁡(1213))\tan\left(\arcsin\left(\frac{12}{13}\right)\right)?

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Cevap: 2.4

Cevap

The exact decimal value of tan⁡(arcsin⁡(1213))\tan\left(\arcsin\left(\frac{12}{13}\right)\right) is 2.42.4.
Letting θ=arcsin⁡(1213)\theta = \arcsin\left(\frac{12}{13}\right), we establish a right triangle in Quadrant I with opposite side length 1212 and hypotenuse length 1313. The adjacent side length is calculated via the Pythagorean theorem as 132−122=5\sqrt{13^2 - 12^2} = 5. The tangent of this angle is the ratio of the opposite side to the adjacent side, 125\frac{12}{5}, which equals 2.42.4.

Adım Adım Çözüm

1
Define the angle using the inverse trigonometric expression
Let θ=arcsin⁡(1213)\theta = \arcsin\left(\frac{12}{13}\right), which implies sin⁡(θ)=1213\sin(\theta) = \frac{12}{13} in Quadrant I.
The inverse sine function returns an angle whose sine is the given ratio, restricted to the principal interval [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
2
Determine cos⁡(θ)\cos(\theta) using the fundamental trigonometric identity
\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13}
Since θ\theta is in Quadrant I, cos⁡(θ)\cos(\theta) is positive.
3
Evaluate tan⁡(θ)\tan(\theta) as the ratio of sine to cosine
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{12/13}{5/13} = \frac{12}{5} = 2.4
Dividing opposite by adjacent (or sine by cosine) yields the exact decimal value 2.42.4.

Anahtar Kavram

Evaluating algebraic values of composite inverse trigonometric expressions