Trigonometry

112 questions

Question 21Question

What is the period of the function f(x)=4cos(13x)f(x) = 4\cos\left(\frac{1}{3}x\right)?

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Answer: 6π6\pi

Answer

The period of the function is 6π6\pi.
The standard period of the cosine function y=cos(x)y = \cos(x) is 2π2\pi. For a transformed trigonometric function of the form y=Acos(Bx)y = A\cos(Bx), the period is given by the formula 2πB\frac{2\pi}{|B|}. In the function f(x)=4cos(13x)f(x) = 4\cos\left(\frac{1}{3}x\right), the coefficient of xx is B=13B = \frac{1}{3}. Dividing the standard period 2π2\pi by 13\frac{1}{3} yields a period of 2π×3=6π2\pi \times 3 = 6\pi.

Step-by-Step Solution

1
Identify the standard period of the parent cosine function and the coefficient BB of the variable xx in the given function.
The parent function is y=cos(x)y = \cos(x), which has a standard period of 2π2\pi. In the function f(x)=4cos(13x)f(x) = 4\cos\left(\frac{1}{3}x\right), the coefficient of xx is B=13B = \frac{1}{3}.
This sets up the parameters needed for the period formula Period=2πB\text{Period} = \frac{2\pi}{|B|}.
2
Substitute the value of BB into the period formula and simplify.
The period is 2π13=2π×3=6π\frac{2\pi}{\frac{1}{3}} = 2\pi \times 3 = 6\pi.
Dividing by a fraction is equivalent to multiplying by its reciprocal, which gives the final horizontal distance for one complete cycle.

Key Concept

Graphs of Trigonometric Functions
Question 22Question

An angle θ\theta in standard position is rotated counterclockwise by 225225^\circ. The terminal ray of the resulting angle lies in the fourth quadrant along the line y=x3y = -x\sqrt{3}. If the radian measure of the smallest positive angle θ\theta is written in simplest form as aπb\frac{a\pi}{b}, where aa and bb are positive integers, what is the value of a+ba + b?

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Answer: 17

Answer

The value of a+ba + b is 1717.
The resulting angle α\alpha lies along the line y=x3y = -x\sqrt{3} in the fourth quadrant, meaning its measure is 300300^\circ plus any multiple of 360360^\circ. Subtracting the counterclockwise rotation of 225225^\circ gives the original angle θ=75\theta = 75^\circ (for the smallest positive angle). Converting 7575^\circ to radians by multiplying by π180\frac{\pi}{180^\circ} yields 5π12\frac{5\pi}{12}. Since the fraction is in simplest form, a=5a = 5 and b=12b = 12, and their sum is 1717.

Step-by-Step Solution

1
Find the angle of the terminal ray after rotation from its equation and quadrant
The terminal ray after rotation is at an angle of 300300^\circ (or 5π3\frac{5\pi}{3} radians)
The line y=x3y = -x\sqrt{3} has a slope of 3-\sqrt{3}, so the angle α\alpha in Quadrant IV satisfies tanα=3\tan\alpha = -\sqrt{3}, which means α=300\alpha = 300^\circ.
2
Set up and solve the equation for the original angle θ\theta before the counterclockwise rotation of 225225^\circ
θ=75+360k\theta = 75^\circ + 360^\circ k
Since the angle was rotated counterclockwise by 225225^\circ to reach the final position of 300300^\circ, we have θ+225=300+360k\theta + 225^\circ = 300^\circ + 360^\circ k.
3
Determine the smallest positive value of θ\theta
θ=75\theta = 75^\circ
Setting k=0k = 0 gives the smallest positive angle of 7575^\circ.
4
Convert the angle θ\theta from degrees to radians
θ=5π12\theta = \frac{5\pi}{12} radians
To convert degrees to radians, multiply by π180\frac{\pi}{180^\circ}, yielding 75π180=5π12\frac{75\pi}{180} = \frac{5\pi}{12}.
5
Calculate the sum of the numerator and denominator of the simplified radian fraction
a+b=17a + b = 17
The fraction 512\frac{5}{12} is in simplest form, so a=5a = 5 and b=12b = 12. The sum is 5+12=175 + 12 = 17.

Key Concept

Converting degree measures to radian measures and finding coterminal angles on the unit circle
Question 23Question

In right triangle PQRPQR, the right angle is at vertex QQ. If the side lengths are PQ=15PQ = 15 and QR=8QR = 8, what is the value of tan(P)\tan(P)?

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Answer: 815\frac{8}{15}

Answer

The tangent of angle PP is 815\frac{8}{15}.
The tangent of angle PP is the ratio of the opposite side (QR=8QR = 8) to the adjacent side (PQ=15PQ = 15). This gives the value 815\frac{8}{15}.

Step-by-Step Solution

1
Identify the reference angle and the sides of the right triangle relative to it.
The reference angle is PP. The side opposite to angle PP is QRQR with a length of 88. The side adjacent to angle PP is PQPQ with a length of 1515. The hypotenuse is PRPR.
To calculate a trigonometric ratio, we must first determine which sides are opposite, adjacent, and the hypotenuse relative to the target angle.
2
Recall the definition of the tangent ratio in a right triangle.
The tangent of an angle is defined as the ratio of the length of the opposite side to the length of the adjacent side: tan(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}.
The question asks for the tangent of angle PP.
3
Substitute the identified side lengths into the tangent formula.
tan(P)=QRPQ=815\tan(P) = \frac{QR}{PQ} = \frac{8}{15}.
Plugging the lengths of the opposite side (88) and the adjacent side (1515) into the tangent ratio yields the final value.

Key Concept

Right triangle trigonometry ratios (SOHCAHTOA), specifically the tangent ratio definition.
Estimated Time:45s
Question 24Question

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

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Answer: -2

Answer

The value of cscθ+cotθ\csc\theta + \cot\theta is 2-2.
By using the difference of squares on the identity sec2θtan2θ=1\sec^2\theta - \tan^2\theta = 1, we get (secθtanθ)(secθ+tanθ)=1(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1. Substituting secθtanθ=3\sec\theta - \tan\theta = 3 gives secθ+tanθ=13\sec\theta + \tan\theta = \frac{1}{3}. Solving the system of equations gives secθ=53\sec\theta = \frac{5}{3} and tanθ=43\tan\theta = -\frac{4}{3}. Since θ\theta lies in Quadrant IV, cosθ=35\cos\theta = \frac{3}{5} and sinθ=45\sin\theta = -\frac{4}{5}. We then find cscθ=54\csc\theta = -\frac{5}{4} and cotθ=34\cot\theta = -\frac{3}{4}, which sum to 2-2.

Step-by-Step Solution

1
Use the Pythagorean identity sec2θtan2θ=1\sec^2\theta - \tan^2\theta = 1, which factors into (secθtanθ)(secθ+tanθ)=1(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1.
Since secθtanθ=3\sec\theta - \tan\theta = 3, we have 3(secθ+tanθ)=1    secθ+tanθ=133(\sec\theta + \tan\theta) = 1 \implies \sec\theta + \tan\theta = \frac{1}{3}.
To establish a second linear equation in terms of secθ\sec\theta and tanθ\tan\theta.
2
Add and subtract the two equations: secθtanθ=3\sec\theta - \tan\theta = 3 and secθ+tanθ=13\sec\theta + \tan\theta = \frac{1}{3}.
Adding them gives 2secθ=103    secθ=532\sec\theta = \frac{10}{3} \implies \sec\theta = \frac{5}{3}. Subtracting the first from the second gives 2tanθ=83    tanθ=432\tan\theta = -\frac{8}{3} \implies \tan\theta = -\frac{4}{3}.
To isolate the values of secθ\sec\theta and tanθ\tan\theta.
3
Find cosθ\cos\theta and sinθ\sin\theta using cosθ=1secθ\cos\theta = \frac{1}{\sec\theta} and sinθ=tanθcosθ\sin\theta = \tan\theta\cos\theta.
cosθ=35\cos\theta = \frac{3}{5} and sinθ=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 cscθ\csc\theta and cotθ\cot\theta using reciprocal identities.
cscθ=1sinθ=54\csc\theta = \frac{1}{\sin\theta} = -\frac{5}{4} and cotθ=1tanθ=34\cot\theta = \frac{1}{\tan\theta} = -\frac{3}{4}.
To obtain the terms of the required sum.
5
Sum the values of cscθ\csc\theta and cotθ\cot\theta.
cscθ+cotθ=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.

Key Concept

Pythagorean and Reciprocal Trigonometric Identities
Question 25Question

For an angle θ\theta such that π<θ<3π2\pi < \theta < \frac{3\pi}{2}, the equation 1sinθcosθ+cosθ1sinθ=103\frac{1 - \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = -\frac{10}{3} is satisfied. What is the value of sinθ\sin\theta?

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Answer: 45-\frac{4}{5}

Answer

The value of sinθ\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(1sinθ)2(1-\sin\theta) using the Pythagorean identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1. The term (1sinθ)(1-\sin\theta) cancels out, resulting in 2cosθ=103\frac{2}{\cos\theta} = -\frac{10}{3}, which implies cosθ=35\cos\theta = -\frac{3}{5}. In Quadrant III, sine is negative, so using the identity sinθ=1cos2θ\sin\theta = -\sqrt{1 - \cos^2\theta} yields the correct value.

Step-by-Step Solution

1
Find a common denominator to add the fractions on the left-hand side of the equation.
The expression becomes (1sinθ)2+cos2θcosθ(1sinθ)\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 sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1.
The numerator simplifies to 12sinθ+sin2θ+cos2θ=12sinθ+1=22sinθ=2(1sinθ)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 (1sinθ)(1 - \sin\theta) from the numerator and denominator, and equate the simplified term to 103-\frac{10}{3}.
The equation simplifies to 2cosθ=103\frac{2}{\cos\theta} = -\frac{10}{3}, which gives cosθ=35\cos\theta = -\frac{3}{5}.
To solve for the cosine of the angle θ\theta.
4
Use the Pythagorean identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1 to calculate sinθ\sin\theta, keeping the quadrant sign rules in mind.
Since π<θ<3π2\pi < \theta < \frac{3\pi}{2} (Quadrant III), sinθ\sin\theta is negative. Therefore, sinθ=1cos2θ=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.

Key Concept

Fundamental Trigonometric Identities
Question 26Question

Match each of the degree measures of angles in standard position on the left with its mathematically equivalent radian measure on the right. Which radian measure corresponds to each degree measure?

Click a left item, then click its matching right item

Items

135-135^\circ
480480^\circ
300-300^\circ
585585^\circ

Matches

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Answer

The correct pairings are: 135-135^\circ matches with 3π4-\frac{3\pi}{4} radians; 480480^\circ matches with 8π3\frac{8\pi}{3} radians; 300-300^\circ matches with 5π3-\frac{5\pi}{3} radians; and 585585^\circ matches with 13π4\frac{13\pi}{4} radians.
Each degree measure is multiplied by π180\frac{\pi}{180^\circ} and simplified to find its equivalent radian measure. This process yields the unique matching pairs: 135-135^\circ to 3π4-\frac{3\pi}{4} radians, 480480^\circ to 8π3\frac{8\pi}{3} radians, 300-300^\circ to 5π3-\frac{5\pi}{3} radians, and 585585^\circ to 13π4\frac{13\pi}{4} radians.

Step-by-Step Solution

1
Recall the formula to convert degrees to radians.
Radian measure = Degree measure ×π180\times \frac{\pi}{180^\circ}
Since a full circle is 360360^\circ or 2π2\pi radians, the conversion ratio simplifies to π\pi radians per 180180^\circ.
2
Convert the first degree measure, 135-135^\circ, to radians.
135×π180=135π180=3π4-135^\circ \times \frac{\pi}{180^\circ} = -\frac{135\pi}{180} = -\frac{3\pi}{4} radians
Dividing the numerator and denominator by their greatest common divisor, 4545, simplifies the fraction to 34-\frac{3}{4}.
3
Convert the second degree measure, 480480^\circ, to radians.
480×π180=480π180=8π3480^\circ \times \frac{\pi}{180^\circ} = \frac{480\pi}{180} = \frac{8\pi}{3} radians
Dividing the numerator and denominator by their greatest common divisor, 6060, simplifies the fraction to 83\frac{8}{3}.
4
Convert the third degree measure, 300-300^\circ, to radians.
300×π180=300π180=5π3-300^\circ \times \frac{\pi}{180^\circ} = -\frac{300\pi}{180} = -\frac{5\pi}{3} radians
Dividing the numerator and denominator by their greatest common divisor, 6060, simplifies the fraction to 53-\frac{5}{3}.
5
Convert the fourth degree measure, 585585^\circ, to radians.
585×π180=585π180=13π4585^\circ \times \frac{\pi}{180^\circ} = \frac{585\pi}{180} = \frac{13\pi}{4} radians
Dividing the numerator and denominator by their greatest common divisor, 4545, simplifies the fraction to 134\frac{13}{4}.

Key Concept

Converting degree measures to equivalent radian measures using the conversion factor π180\frac{\pi}{180^\circ}.

Alternative Method

Alternatively, you can recall key benchmark angles on the unit circle (such as 45=π445^\circ = \frac{\pi}{4} radians and 60=π360^\circ = \frac{\pi}{3} radians) and express each angle as an integer multiple of these benchmarks. For example, 135-135^\circ is 3×45-3 \times 45^\circ, which corresponds to 3×π4=3π4-3 \times \frac{\pi}{4} = -\frac{3\pi}{4} radians.
Estimated Time:1m 30s
Question 27Question

What is the period of the function f(x)=tan(3x)f(x) = \tan(3x)?

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Answer: π3\frac{\pi}{3}

Answer

The period of the function is π3\frac{\pi}{3}.
The parent function y=tan(x)y = \tan(x) has a standard period of π\pi. To find the period of a transformed tangent function of the form y=tan(Bx)y = \tan(Bx), the standard period must be divided by the coefficient of xx, yielding πB\frac{\pi}{|B|}. Substituting B=3B = 3 gives π3\frac{\pi}{3}.

Step-by-Step Solution

1
Determine the standard period of the parent function.
The parent function is the tangent function, y=tan(x)y = \tan(x), which has a standard period of π\pi.
The tangent function completes one full cycle of its graph between π2-\frac{\pi}{2} and π2\frac{\pi}{2}.
2
Identify the horizontal compression/stretch coefficient from the given equation.
In f(x)=tan(3x)f(x) = \tan(3x), the coefficient of xx is B=3B = 3.
This coefficient determines how many cycles occur in a standard interval.
3
Calculate the period using the formula for the tangent function.
Period = πB=π3\frac{\pi}{|B|} = \frac{\pi}{3}.
Dividing the standard period of π\pi by the absolute value of the coefficient BB gives the compressed period of the transformed function.

Key Concept

The period of a transformed tangent function y=tan(Bx)y = \tan(Bx) is given by πB\frac{\pi}{|B|}.
Question 28Question

A 1010-foot ladder leans against a vertical wall. The base of the ladder is 66 feet away from the bottom of the wall, and the top of the ladder reaches a height of 88 feet up the wall. If θ\theta is the angle formed between the ladder and the ground, what is the value of cos(θ)\cos(\theta)?

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Answer: 35\frac{3}{5}

Answer

The correct answer is 35\frac{3}{5}, which represents the ratio of the adjacent side (66 feet) to the hypotenuse (1010 feet) for the angle θ\theta formed between the ladder and the ground.
The correct answer is the value 35\frac{3}{5}. The angle θ\theta is formed between the ladder (hypotenuse) and the ground (adjacent side). The cosine of an angle in a right triangle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse. Here, the adjacent side is 66 feet and the hypotenuse is 1010 feet. Therefore, cos(θ)=610\cos(\theta) = \frac{6}{10}, which simplifies to 35\frac{3}{5}.

Step-by-Step Solution

1
Identify the parts of the right triangle relative to the angle θ\theta formed between the ladder and the ground.
The hypotenuse is the length of the ladder (1010 feet). The side adjacent to θ\theta is the distance along the ground from the wall to the base of the ladder (66 feet). The side opposite to θ\theta is the height up the wall (88 feet).
To apply trigonometric ratios, we must first map the given dimensions of the scenario to the sides of a right triangle relative to the reference angle.
2
Recall the definition of the cosine ratio in a right triangle.
cos(θ)=AdjacentHypotenuse\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}
We need to identify the mathematical definition of cosine to set up the calculation.
3
Substitute the identified side lengths into the cosine ratio and simplify the fraction.
cos(θ)=610=35\cos(\theta) = \frac{6}{10} = \frac{3}{5}
Substituting the values of 66 feet for the adjacent side and 1010 feet for the hypotenuse gives 610\frac{6}{10}, which simplifies to 35\frac{3}{5} when both the numerator and denominator are divided by their greatest common divisor, 22.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)
Question 29Question

An angle θ\theta lies in the third quadrant, where π<θ<3π2\pi < \theta < \frac{3\pi}{2}. If sinθcosθ=15\sin\theta - \cos\theta = -\frac{1}{5}, what is the value of the expression 125(sin3θ+cos3θ)125(\sin^3\theta + \cos^3\theta)?

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Answer: -91

Answer

The correct value of the expression is -91.
Squaring the equation sinθcosθ=15\sin\theta - \cos\theta = -\frac{1}{5} gives 12sinθcosθ=1251 - 2\sin\theta\cos\theta = \frac{1}{25}, which simplifies to sinθcosθ=1225\sin\theta\cos\theta = \frac{12}{25}. We then find the square of the sum: (sinθ+cosθ)2=1+2sinθcosθ=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 sinθ+cosθ=75\sin\theta + \cos\theta = -\frac{7}{5}. Factoring the sum of cubes gives sin3θ+cos3θ=(sinθ+cosθ)(sin2θsinθcosθ+cos2θ)=(75)(11225)=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.

Step-by-Step Solution

1
Square both sides of the equation sinθcosθ=15\sin\theta - \cos\theta = -\frac{1}{5}
sin2θ2sinθcosθ+cos2θ=125    12sinθcosθ=125    sinθcosθ=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 sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1.
2
Determine the value of sinθ+cosθ\sin\theta + \cos\theta using the identity (sinθ+cosθ)2=1+2sinθcosθ(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta
(sinθ+cosθ)2=1+2(1225)=4925    sinθ+cosθ=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 sinθ\sin\theta and cosθ\cos\theta must be negative, meaning their sum is also negative.
3
Use the sum of cubes factorization to evaluate sin3θ+cos3θ\sin^3\theta + \cos^3\theta
sin3θ+cos3θ=(sinθ+cosθ)(sin2θsinθcosθ+cos2θ)=(75)(11225)=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.

Key Concept

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

Alternative Method

Instead of applying algebraic identities to find the sum of cubes directly, we can solve for the individual values of sinθ\sin\theta and cosθ\cos\theta from the system of equations: sinθcosθ=15\sin\theta - \cos\theta = -\frac{1}{5} and sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1. Substituting sinθ=cosθ15\sin\theta = \cos\theta - \frac{1}{5} into the second equation yields 2cos2θ25cosθ2425=02\cos^2\theta - \frac{2}{5}\cos\theta - \frac{24}{25} = 0, which factors as (5cosθ+3)(5cosθ4)=0(5\cos\theta + 3)(5\cos\theta - 4) = 0. Since θ\theta is in Quadrant III, cosθ=35\cos\theta = -\frac{3}{5} and sinθ=45\sin\theta = -\frac{4}{5}. Evaluating 125(sin3θ+cos3θ)125(\sin^3\theta + \cos^3\theta) directly with these values gives 125((45)3+(35)3)=125(6412527125)=91125((-\frac{4}{5})^3 + (-\frac{3}{5})^3) = 125(-\frac{64}{125} - \frac{27}{125}) = -91.
Estimated Time:2m 30s
Question 30Question

The graph of the function f(x)=acos(b(xc))+df(x) = a \cos(b(x - c)) + d is shown below for constants a>0a > 0, b>0b > 0, c[0,π]c \in [0, \pi], and dd. The graph has a local maximum at (π3,5)(\frac{\pi}{3}, 5) and the nearest local minimum to its right is at (5π6,1)(\frac{5\pi}{6}, -1). What is the y-intercept of the graph of f(x)f(x)?

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Answer: 12\frac{1}{2}

Answer

The y-intercept of the graph is 12\frac{1}{2}
The correct answer is 12\frac{1}{2}. The amplitude of the function is a=5(1)2=3a = \frac{5 - (-1)}{2} = 3, and the midline is d=5+(1)2=2d = \frac{5 + (-1)}{2} = 2. The distance between the consecutive local maximum and local minimum represents half of a period: T2=5π6π3=π2\frac{T}{2} = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{\pi}{2}, meaning the period is T=πT = \pi, so b=2ππ=2b = \frac{2\pi}{\pi} = 2. A local maximum occurs when the argument of the cosine function is 00, so 2(π3c)=0    c=π32(\frac{\pi}{3} - c) = 0 \implies c = \frac{\pi}{3}. This gives the equation f(x)=3cos(2(xπ3))+2f(x) = 3 \cos(2(x - \frac{\pi}{3})) + 2. Finding the y-intercept requires evaluating the function at x=0x = 0: f(0)=3cos(2(0π3))+2=3cos(2π3)+2=3(12)+2=12f(0) = 3 \cos(2(0 - \frac{\pi}{3})) + 2 = 3 \cos(-\frac{2\pi}{3}) + 2 = 3(-\frac{1}{2}) + 2 = \frac{1}{2}.

Step-by-Step Solution

1
Find the amplitude aa and midline dd of the trigonometric function.
a=3a = 3 and d=2d = 2
The amplitude is half the distance between the maximum and minimum values: a=5(1)2=3a = \frac{5 - (-1)}{2} = 3. The midline is the average of these values: d=5+(1)2=2d = \frac{5 + (-1)}{2} = 2.
2
Determine the period TT and the frequency coefficient bb.
T=πT = \pi and b=2b = 2
The horizontal distance between a consecutive maximum and minimum is half of the period: 5π6π3=π2\frac{5\pi}{6} - \frac{\pi}{3} = \frac{\pi}{2}. Thus, the full period is T=πT = \pi. Since T=2πbT = \frac{2\pi}{b}, we find b=2b = 2.
3
Determine the horizontal phase shift cc.
c=π3c = \frac{\pi}{3}
A cosine function achieves its maximum when its argument is a multiple of 2π2\pi. Since the maximum is at x=π3x = \frac{\pi}{3}, we set 2(π3c)=02(\frac{\pi}{3} - c) = 0, giving c=π3c = \frac{\pi}{3}.
4
Evaluate the function at x=0x = 0 to find the y-intercept.
f(0)=12f(0) = \frac{1}{2}
Substitute the parameters into the function: f(x)=3cos(2(xπ3))+2f(x) = 3 \cos(2(x - \frac{\pi}{3})) + 2. Substituting x=0x = 0 gives f(0)=3cos(2π3)+2=3(12)+2=12f(0) = 3 \cos(-\frac{2\pi}{3}) + 2 = 3(-\frac{1}{2}) + 2 = \frac{1}{2}.

Key Concept

Determining the equation of a transformed trigonometric function from key features (maximum and minimum points) and evaluating it.
Estimated Time:3m 0s
Question 31Question

In right triangle ABCABC, the right angle is at vertex BB, AB=12AB = 12, and BC=5BC = 5. A line segment BDBD is drawn perpendicular to the hypotenuse ACAC such that DD lies on ACAC. From point DD, a perpendicular line segment DEDE is drawn to side ABAB, where EE lies on ABAB. What is the length of segment DEDE?

Show answer & explanation

Answer: 720169\frac{720}{169}

Answer

720169\frac{720}{169}
The correct answer is 720169\frac{720}{169}. First, find the hypotenuse of the right triangle ABCABC using the Pythagorean theorem: AC=AB2+BC2=122+52=13AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 5^2} = 13. In right triangle ABCABC, the sine of angle AA is sin(A)=BCAC=513\sin(\angle A) = \frac{BC}{AC} = \frac{5}{13}. Next, in right triangle ABDABD (which has a right angle at DD), the length of the altitude BDBD can be found using the sine of angle AA: BD=ABsin(A)=12513=6013BD = AB \sin(\angle A) = 12 \cdot \frac{5}{13} = \frac{60}{13}. In right triangle BDEBDE (which has a right angle at EE), the angle BDE\angle BDE is equal to A\angle A because both are complementary to ABD\angle ABD (or EBD\angle EBD). Therefore, the adjacent side DEDE is found using the cosine of angle BDE\angle BDE: DE=BDcos(BDE)=BDcos(A)=60131213=720169DE = BD \cos(\angle BDE) = BD \cos(\angle A) = \frac{60}{13} \cdot \frac{12}{13} = \frac{720}{169}.

Step-by-Step Solution

1
Find the hypotenuse ACAC of the right triangle ABCABC using the Pythagorean theorem.
AC=13AC = 13
The hypotenuse is needed to find the trigonometric ratios of angle AA.
2
Determine sin(A)\sin(\angle A) and cos(A)\cos(\angle A) from right triangle ABCABC.
sin(A)=513\sin(\angle A) = \frac{5}{13} and cos(A)=1213\cos(\angle A) = \frac{12}{13}
These trigonometric ratios are needed for the calculations in the nested right triangles.
3
Find the length of altitude BDBD in right triangle ABDABD using sin(A)\sin(\angle A).
BD=6013BD = \frac{60}{13}
BDBD serves as the hypotenuse for the next right triangle BDEBDE.
4
Identify that BDE=A\angle BDE = \angle A and calculate the length of DEDE in right triangle BDEBDE.
DE=720169DE = \frac{720}{169}
Since BDE=A\angle BDE = \angle A, DE=BDcos(BDE)=BDcos(A)=60131213=720169DE = BD \cos(\angle BDE) = BD \cos(\angle A) = \frac{60}{13} \cdot \frac{12}{13} = \frac{720}{169}.

Key Concept

Applying SOHCAHTOA (sine and cosine definitions) sequentially across nested right triangles by identifying equal angles.
Estimated Time:2m 0s
Question 32Question

In triangle ABCABC, the length of side aa is 55 centimeters, the length of side bb is 88 centimeters, and the measure of angle CC is 6060^\circ. What is the length, in centimeters, of side cc?

Show answer & explanation

Answer: 7

Answer

The length of side cc is 77 centimeters.
Applying the Law of Cosines directly to the given Side-Angle-Side (SAS) triangle yields c2=52+822(5)(8)cos60=25+6440=49c^2 = 5^2 + 8^2 - 2(5)(8) \cos 60^\circ = 25 + 64 - 40 = 49, which gives c=7c = 7.

Step-by-Step Solution

1
Identify the given values and the appropriate formula.
We are given two sides, a=5a = 5 and b=8b = 8, and the included angle C=60C = 60^\circ. To find the opposite side cc, we use the Law of Cosines: c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab \cos C.
The Law of Cosines relates three sides of a triangle to the cosine of one of its angles, which is applicable for Side-Angle-Side (SAS) configurations.
2
Substitute the known values into the Law of Cosines equation.
c2=52+822(5)(8)cos60c^2 = 5^2 + 8^2 - 2(5)(8) \cos 60^\circ
Plugging the given values into the formula allows us to solve for the unknown side cc.
3
Evaluate the trigonometric and arithmetic terms.
Since cos60=0.5\cos 60^\circ = 0.5, we get:
c2=25+6480(0.5)c^2 = 25 + 64 - 80(0.5)
c2=8940c^2 = 89 - 40
c2=49c^2 = 49
Simplifying the expression step-by-step leads to the value of c2c^2.
4
Take the square root of both sides to find the side length.
c=49=7c = \sqrt{49} = 7
Since side lengths must be positive, the square root of 4949 gives the exact length of side cc.

Key Concept

Law of Cosines
Question 33Question

A coordinate grid contains a right triangle, XYZXYZ, with the right angle located at vertex ZZ. The horizontal leg XZXZ has a length of 2424 units, and the vertical leg YZYZ has a length of 77 units. What is the value of cos(X)\cos(X)?

Show answer & explanation

Answer: 2425\frac{24}{25}

Answer

The correct answer is the option representing 2425\frac{24}{25}.
The cosine of an angle in a right triangle is the ratio of the length of the adjacent leg to the length of the hypotenuse. The hypotenuse of the triangle is calculated as 2525 units using the Pythagorean theorem. Relative to angle XX, the adjacent leg is XZ=24XZ = 24 and the hypotenuse is XY=25XY = 25. Therefore, cos(X)=2425\cos(X) = \frac{24}{25}.

Step-by-Step Solution

1
Use the Pythagorean theorem to calculate the hypotenuse XYXY.
XY=XZ2+YZ2=242+72=576+49=625=25XY = \sqrt{XZ^2 + YZ^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25.
To calculate any primary trigonometric ratio, we need the lengths of the relevant sides, including the hypotenuse.
2
Identify the adjacent side relative to angle XX.
The leg adjacent to angle XX is XZ=24XZ = 24.
Angle XX is formed by the hypotenuse XYXY and the adjacent leg XZXZ.
3
Apply the definition of cosine (adjacent over hypotenuse).
cos(X)=XZXY=2425\cos(X) = \frac{XZ}{XY} = \frac{24}{25}.
By SOHCAHTOA, cosine is the ratio of the adjacent side length to the hypotenuse length.

Key Concept

Using SOHCAHTOA definitions to find trigonometric ratios in a right triangle.
Estimated Time:1m 0s
Question 34Question

For an angle θ\theta such that π2<θ<π\frac{\pi}{2} < \theta < \pi, if cosθ1sinθ=3\frac{\cos\theta}{1 - \sin\theta} = -3, what is the value of sinθtanθ\sin\theta - \tan\theta?

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Answer: 3215\frac{32}{15}

Answer

3215\frac{32}{15}
By multiplying the numerator and denominator of the given expression by 1+sinθ1 + \sin\theta and utilizing the Pythagorean identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1, we find that secθ+tanθ=3\sec\theta + \tan\theta = -3. Since sec2θtan2θ=1\sec^2\theta - \tan^2\theta = 1, it follows that secθtanθ=13\sec\theta - \tan\theta = -\frac{1}{3}. Solving this system of equations yields secθ=53\sec\theta = -\frac{5}{3} (which means cosθ=35\cos\theta = -\frac{3}{5}) and tanθ=43\tan\theta = -\frac{4}{3}. In Quadrant II, sine is positive, so sinθ=tanθcosθ=45\sin\theta = \tan\theta \cos\theta = \frac{4}{5}. Thus, sinθtanθ=45(43)=12+2015=3215\sin\theta - \tan\theta = \frac{4}{5} - \left(-\frac{4}{3}\right) = \frac{12 + 20}{15} = \frac{32}{15}.

Step-by-Step Solution

1
Multiply the numerator and denominator of the given expression by 1+sinθ1 + \sin\theta to simplify it.
cosθ(1+sinθ)(1sinθ)(1+sinθ)=3    cosθ(1+sinθ)1sin2θ=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 1sin2θ=cos2θ1 - \sin^2\theta = \cos^2\theta into the denominator.
cosθ(1+sinθ)cos2θ=1+sinθcosθ=secθ+tanθ=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 sec2θtan2θ=1\sec^2\theta - \tan^2\theta = 1 to find the difference of secant and tangent.
secθtanθ=1secθ+tanθ=13=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 secθ\sec\theta and tanθ\tan\theta.
secθ=53\sec\theta = -\frac{5}{3} and tanθ=43\tan\theta = -\frac{4}{3}
Adding the two equations yields 2secθ=103    secθ=532\sec\theta = -\frac{10}{3} \implies \sec\theta = -\frac{5}{3}, and subtracting them yields 2tanθ=83    tanθ=432\tan\theta = -\frac{8}{3} \implies \tan\theta = -\frac{4}{3}.
5
Find cosθ\cos\theta and sinθ\sin\theta, and calculate the final value of sinθtanθ\sin\theta - \tan\theta.
cosθ=35\cos\theta = -\frac{3}{5}, sinθ=45\sin\theta = \frac{4}{5}, and sinθtanθ=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 sinθ=tanθcosθ=(43)(35)=45\sin\theta = \tan\theta \cdot \cos\theta = \left(-\frac{4}{3}\right)\left(-\frac{3}{5}\right) = \frac{4}{5}.

Key Concept

Fundamental Trigonometric Identities
Estimated Time:2m 0s
Question 35Question

A straight skateboard ramp has a vertical height of 99 feet and a horizontal length of 1212 feet along the ground. The vertical height and horizontal length meet at a right angle. What is the sine of the angle that the ramp makes with the ground?

Show answer & explanation

Answer: 35\frac{3}{5}

Answer

The value of the sine of the angle is 35\frac{3}{5}.
The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. The opposite side is the vertical height of 99 feet. Using the Pythagorean theorem, the hypotenuse (the length of the ramp) is calculated as 92+122=15\sqrt{9^2 + 12^2} = 15 feet. Therefore, the sine of the angle is 915\frac{9}{15}, which simplifies to 35\frac{3}{5}.

Step-by-Step Solution

1
Identify the lengths of the legs of the right triangle and calculate the length of the hypotenuse using the Pythagorean theorem.
The hypotenuse (ramp length) is 1515 feet.
The vertical height (99 feet) and horizontal length (1212 feet) form the perpendicular legs of a right triangle. The length of the ramp is the hypotenuse: 92+122=81+144=225=15\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 feet.
2
Identify the opposite side and hypotenuse relative to the angle the ramp makes with the ground, then apply the sine ratio definition.
The sine of the angle is 35\frac{3}{5}.
The angle the ramp makes with the ground is at the bottom vertex. The side opposite this angle is the vertical height (99 feet), and the hypotenuse is the ramp length (1515 feet). The sine ratio is sin(θ)=oppositehypotenuse=915\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{15}, which simplifies to 35\frac{3}{5}.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)
Estimated Time:1m 0s
Question 36Question

A point on the unit circle starts at (1,0)(1, 0) and undergoes a counterclockwise rotation of 570570^\circ about the origin. What is the radian measure of the angle in the interval [0,2π)[0, 2\pi) that corresponds to the point's final position?

Show answer & explanation

Answer: 7π6\frac{7\pi}{6}

Answer

The correct answer is the option representing 7π6\frac{7\pi}{6} radians.
The correct answer represents 7π6\frac{7\pi}{6} radians. To find the position on the unit circle after a counterclockwise rotation of 570570^\circ, we first determine the coterminal angle within one full rotation (360360^\circ). Subtracting 360360^\circ from 570570^\circ gives 210210^\circ. We then convert this angle to radians by multiplying by π180\frac{\pi}{180^\circ}, which simplifies to 7π6\frac{7\pi}{6} radians.

Step-by-Step Solution

1
Find a coterminal angle for 570570^\circ within the standard [0,360)[0^\circ, 360^\circ) range.
570360=210570^\circ - 360^\circ = 210^\circ
Since a full rotation is 360360^\circ, subtracting 360360^\circ yields an angle in the same position on the unit circle but within one full rotation.
2
Convert the coterminal angle from degrees to radians by multiplying by π180\frac{\pi}{180^\circ}.
210×π180=210π180=7π6210^\circ \times \frac{\pi}{180^\circ} = \frac{210\pi}{180} = \frac{7\pi}{6} radians
To convert degrees to radians, multiply by the conversion factor π180\frac{\pi}{180^\circ}.
3
Verify that the resulting angle is in the interval [0,2π)[0, 2\pi).
Since 07π6<2π0 \le \frac{7\pi}{6} < 2\pi, the angle is in the correct interval.
The question requires the final angle to be in the interval [0,2π)[0, 2\pi).

Key Concept

Finding coterminal angles and converting degree measures to radian measures on the unit circle.

Alternative Method

Convert the rotation to radians first: 570×π180=19π6570^\circ \times \frac{\pi}{180^\circ} = \frac{19\pi}{6} radians. Then, subtract 2π2\pi to find the coterminal angle: 19π62π=7π6\frac{19\pi}{6} - 2\pi = \frac{7\pi}{6} radians.
Estimated Time:1m 15s
Question 37Question

An angle θ\theta in standard position measures 750-750^\circ. Let ϕ\phi be the coterminal angle of θ\theta such that 0ϕ<2π0 \le \phi < 2\pi radians. If ϕ=aπb\phi = \frac{a\pi}{b}, where aa and bb are positive integers with no common factors, what is the value of a+ba + b?

Show answer & explanation

Answer: 17

Answer

17
To find the coterminal angle ϕ\phi in the range [0,2π)[0, 2\pi) radians, we first determine the coterminal angle in degrees by adding multiples of 360360^\circ. Adding 3×360=10803 \times 360^\circ = 1080^\circ to 750-750^\circ results in 330330^\circ. We then convert 330330^\circ to radians by multiplying by π180\frac{\pi}{180^\circ}, which simplifies to 11π6\frac{11\pi}{6}. Since 1111 and 66 are positive integers with no common factors, a=11a = 11 and b=6b = 6. The sum a+ba + b is 11+6=1711 + 6 = 17.

Step-by-Step Solution

1
Find the positive coterminal angle of 750-750^\circ within one full rotation.
330330^\circ
Adding 10801080^\circ (three full rotations of 360360^\circ) to 750-750^\circ brings the angle within the standard range of [0,360)[0^\circ, 360^\circ).
2
Convert the angle from degrees to radians.
11π6\frac{11\pi}{6} radians
Multiplying the degree measure by π180\frac{\pi}{180^\circ} and simplifying the fraction converts it to radians.
3
Identify the values of aa and bb and compute their sum.
1717
Comparing 11π6\frac{11\pi}{6} to aπb\frac{a\pi}{b} shows a=11a = 11 and b=6b = 6, which share no common factors. Their sum is 11+6=1711 + 6 = 17.

Key Concept

Finding coterminal angles and converting angle measures between degrees and radians.
Estimated Time:1m 30s
Question 38Question

In triangle PQRPQR, the measure of angle PP is 3030^\circ, the measure of angle QQ is 4545^\circ, and the length of side QRQR is 1010 units. Which of the following expressions represents the length, in units, of side PRPR?

Show answer & explanation

Answer: 10sin(45)sin(30)\frac{10\sin(45^\circ)}{\sin(30^\circ)}

Answer

The length of side PRPR is represented by the expression 10sin(45)sin(30)\frac{10\sin(45^\circ)}{\sin(30^\circ)}.
The correct answer is derived by setting up the Law of Sines proportion: PRsin(45)=10sin(30)\frac{PR}{\sin(45^\circ)} = \frac{10}{\sin(30^\circ)}. Multiplying both sides by sin(45)\sin(45^\circ) yields PR=10sin(45)sin(30)PR = \frac{10\sin(45^\circ)}{\sin(30^\circ)}.

Step-by-Step Solution

1
Identify the known values and corresponding angle-side pairs in triangle PQRPQR.
Angle P=30P = 30^\circ is opposite to side QR=10QR = 10, and angle Q=45Q = 45^\circ is opposite to side PRPR.
This allows us to set up the appropriate trigonometric relationship to solve for the unknown side.
2
Apply the Law of Sines to relate the ratios of the side lengths to the sines of their opposite angles.
PRsin(Q)=QRsin(P)\frac{PR}{\sin(Q)} = \frac{QR}{\sin(P)}, which becomes PRsin(45)=10sin(30)\frac{PR}{\sin(45^\circ)} = \frac{10}{\sin(30^\circ)}.
The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant for all three sides.
3
Isolate the variable representing the length of side PRPR.
PR=10sin(45)sin(30)PR = \frac{10\sin(45^\circ)}{\sin(30^\circ)}.
Multiply both sides of the equation by sin(45)\sin(45^\circ) to solve for PRPR.

Key Concept

The Law of Sines relates the side lengths of a triangle to the sines of its angles: asin(A)=bsin(B)=csin(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}.
Estimated Time:1m 0s
Question 39Question

For the trigonometric equations on the left, which description on the right correctly matches the graphical features of each equation?

Click a left item, then click its matching right item

Items

y=3sin(2xπ3)+1y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1
y=2cos(12x+π4)1y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1
y=tan(3xπ2)+2y = -\tan\left(3x - \frac{\pi}{2}\right) + 2
y=3sin(23x+π)2y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2

Matches

Show answer & explanation

Answer

The correct matches are: the equation y=3sin(2xπ3)+1y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1 matches the description with a period of π\pi, a range of [2,4][-2, 4], and a positive yy-intercept; the equation y=2cos(12x+π4)1y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1 matches the description with a period of 4π4\pi, a range of [3,1][-3, 1], and a phase shift of π2\frac{\pi}{2} units to the left; the equation y=tan(3xπ2)+2y = -\tan\left(3x - \frac{\pi}{2}\right) + 2 matches the description with a period of π3\frac{\pi}{3} that is undefined at the yy-axis; and the equation y=3sin(23x+π)2y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2 matches the description with a period of 3π3\pi, a range of [5,1][-5, 1], and a yy-intercept of (0,2)(0, -2).
Each equation is correctly paired with the unique graphical properties determined by calculating its amplitude, period, range, phase shift, and yy-intercept. The first equation features a period of π\pi, a range of [2,4][-2, 4], and a positive yy-intercept. The second equation has a period of 4π4\pi, a range of [3,1][-3, 1], and a phase shift of π2\frac{\pi}{2} units left. The third equation has a period of π3\frac{\pi}{3} and is undefined at the yy-axis since x=0x=0 creates a vertical asymptote. The fourth equation has a period of 3π3\pi, a range of [5,1][-5, 1], and a yy-intercept of (0,2)(0, -2).

Step-by-Step Solution

1
Determine the period, range, and y-intercept for the function y=3sin(2xπ3)+1y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1.
The period is π\pi, range is [2,4][-2, 4], and yy-intercept is 1+332>01 + \frac{3\sqrt{3}}{2} > 0.
The period is 2π2=π\frac{2\pi}{2} = \pi, the range is [13,1+3]=[2,4][1-3, 1+3] = [-2, 4], and evaluating at x=0x=0 yields y=3sin(π/3)+1>0y = -3\sin(-\pi/3) + 1 > 0.
2
Determine the period, range, and phase shift for the function y=2cos(12x+π4)1y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1.
The period is 4π4\pi, range is [3,1][-3, 1], and phase shift is π2\frac{\pi}{2} units left.
The period is 2π1/2=4π\frac{2\pi}{1/2} = 4\pi, the range is [12,1+2]=[3,1][-1-2, -1+2] = [-3, 1], and factoring the argument yields 12(x+π2)\frac{1}{2}(x + \frac{\pi}{2}), giving a shift of π2\frac{\pi}{2} units left.
3
Determine the period and domain restriction for the function y=tan(3xπ2)+2y = -\tan\left(3x - \frac{\pi}{2}\right) + 2.
The period is π3\frac{\pi}{3} and the function is undefined at x=0x = 0 (the yy-axis).
For tangent, the period is π3\frac{\pi}{3}. At x=0x=0, the argument is π2-\frac{\pi}{2}, where tangent is undefined, meaning the function has a vertical asymptote at the yy-axis.
4
Determine the period, range, and y-intercept for the function y=3sin(23x+π)2y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2.
The period is 3π3\pi, range is [5,1][-5, 1], and yy-intercept is (0,2)(0, -2).
The period is 2π2/3=3π\frac{2\pi}{2/3} = 3\pi, the range is [23,2+3]=[5,1][-2-3, -2+3] = [-5, 1], and at x=0x=0, y=3sin(π)2=2y = 3\sin(\pi) - 2 = -2.

Key Concept

Identifying graphs of trigonometric functions from their equations by determining amplitude, period, phase shift, midline, range, and asymptotes.
Question 40Question

A surveyor stands at point AA on flat ground and measures the angle of elevation to the top of a vertical tower, TT, at point CC to be 3030^\circ. Another surveyor at point BB, which is 100100 meters away from AA on the same flat ground, measures CAB=40\angle CAB = 40^\circ and CBA=65\angle CBA = 65^\circ. Which of the following expressions represents the height, in meters, of the tower?

Show answer & explanation

Answer: 100sin(65)tan(30)sin(75)\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}

Answer

The correct expression is 100sin(65)tan(30)sin(75)\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}
The correct expression is derived by first applying the Law of Sines to find the length of the ground segment ACAC, and then using right-triangle trigonometry to determine the height of the tower. In triangle ABCABC, the third angle ACB\angle ACB is 180(40+65)=75180^\circ - (40^\circ + 65^\circ) = 75^\circ. The Law of Sines gives ACsin(65)=100sin(75)\frac{AC}{\sin(65^\circ)} = \frac{100}{\sin(75^\circ)}, which simplifies to AC=100sin(65)sin(75)AC = \frac{100\sin(65^\circ)}{\sin(75^\circ)}. Since the tower is vertical, triangle ACTACT is a right triangle with tan(30)=hAC\tan(30^\circ) = \frac{h}{AC}. Substituting ACAC yields h=100sin(65)tan(30)sin(75)h = \frac{100\sin(65^\circ)\tan(30^\circ)}{\sin(75^\circ)}.

Step-by-Step Solution

1
Calculate the measure of the third angle in the ground triangle ABC\triangle ABC.
ACB=180(40+65)=75\angle ACB = 180^\circ - (40^\circ + 65^\circ) = 75^\circ
The sum of angles in any triangle must equal 180180^\circ.
2
Apply the Law of Sines to find the distance from point AA to the base of the tower at point CC (ACAC).
ACsin(65)=100sin(75)AC=100sin(65)sin(75)\frac{AC}{\sin(65^\circ)} = \frac{100}{\sin(75^\circ)} \Rightarrow AC = \frac{100 \sin(65^\circ)}{\sin(75^\circ)}
The Law of Sines states that the ratio of a side length to the sine of its opposite angle is constant in a triangle.
3
Use the right-triangle trigonometric ratio for the vertical tower height hh from point AA.
tan(30)=hACh=ACtan(30)\tan(30^\circ) = \frac{h}{AC} \Rightarrow h = AC \tan(30^\circ)
In right triangle ACT\triangle ACT, the tangent of the angle of elevation is the ratio of the opposite side (height hh) to the adjacent side (ACAC).
4
Substitute the expression for ACAC into the equation for hh.
h=100sin(65)tan(30)sin(75)h = \frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}
Replacing ACAC with its equivalent algebraic expression yields the final height in terms of the given parameters.

Key Concept

Applying the Law of Sines to find a missing side length in a non-right triangle and then using right-triangle trigonometric ratios to solve a 3D geometry problem.
Estimated Time:2m 0s
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