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SAT II Math I : Trigonometry

Study concepts, example questions & explanations for SAT II Math I

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All SAT II Math I Resources

6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Understanding Sine, Cosine, And Tangent

If sin(x)=0.5 , what is sin(8x) if is between and $2\pi$ ?

Possible Answers:

0.5

-1.5

$\frac{\sqrt{3}}{2}$

$-\frac{\sqrt{3}}{2}$

-0.5

Correct answer:

$-\frac{\sqrt{3}}{2}$

Explanation:

Recall that $sin(\frac{\pi}{6}) =0.5$ .

Therefore, we are looking for $sin(8*\frac{\pi}{6})$ or $sin(\frac{4\pi}{3})$ .

Now, this has a reference angle of $\frac{\pi}{3}$ , but it is in the third quadrant. This means that the value will be negative. The value of $sin(\frac{\pi}{3})$ is $\frac{\sqrt{3}}{2}$ . However, given the quadrant of our angle, it will be $-\frac{\sqrt{3}}{2}$ .

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Example Question #1 : Sine, Cosine, Tangent

Determine the exact value of 8cos(45) .

Possible Answers:

$4\sqrt2$

$6\sqrt2$

$8\sqrt2$

$\frac{\sqrt2}{8}$

$\frac{\sqrt2}{4}$

Correct answer:

$4\sqrt2$

Explanation:

The exact value of cos(45) is the x-value when the angle is 45 degrees on the unit circle.

The x-value of this angle is $\frac{\sqrt2}{2}$ .

$8cos(45)= 8\left(\frac{\sqrt{2}}{2}\right)=4\sqrt2$

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Example Question #2 : Sine, Cosine, Tangent

Sine

Which of the following is equal to cos(x)?

Possible Answers:

tan(x)

sin(y)

cos(y)

sin(x)

tan(y)

Correct answer:

sin(y)

Explanation:

Remember SOH-CAH-TOA! That means:

$\small cos(x)=\frac{adjacent}{hypotenuse}=\frac{b}{c}$

$\small \small \small sin(x)=\frac{a}{c}$ $\small \small \small \small \small sin(y)=\frac{b}{c}$

$\small \small \small \small cos(y)=\frac{a}{c}$

$\small \small \small \small \small tan(x)=\frac{a}{b}$ $\small \small \small \small tan(y)=\frac{b}{a}$

sin(y) is equal to cos(x)

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Example Question #2 : Sin, Cos, Tan

Find the value of $\frac{1}{2}sin(45)+ tan(60)$ .

Possible Answers:

$\frac{\sqrt2+4\sqrt3}{4}$

$\frac{\sqrt2+4\sqrt3}{2}$

$\frac{3\sqrt2+\sqrt3}{12}$

$\frac{\sqrt2-4\sqrt3}{2}$

$\sqrt2+\sqrt3$

Correct answer:

$\frac{\sqrt2+4\sqrt3}{4}$

Explanation:

To find the value of $\frac{1}{2}sin(45)+ tan(60)$ , solve each term separately.

$\frac{1}{2}sin(45)=\frac{1}{2} \cdot \frac{\sqrt2}{2} = \frac{\sqrt2}{4}$

$tan(60) = \sqrt3$

Sum the two terms.

$\frac{\sqrt2}{4}+\sqrt3 = \frac{\sqrt2}{4}+\frac{4\sqrt3}{4} = \frac{\sqrt2+4\sqrt3}{4}$

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Example Question #11 : Trigonometry

Calculate $tan(\frac{4\pi}{3})$ .

Possible Answers:

$\sqrt{3}$

$-\sqrt{3}$

Correct answer:

$\sqrt{3}$

Explanation:

The tangent function has a period of $\pi$ units. That is,

$tan(x+n\pi)=tanx$

for all $n\in\mathbb{Z}$ .

Since $\frac{4\pi}{3}=\frac{\pi}{3}+\pi$ , we can rewrite the original expression $tan(\frac{4\pi}{3})$ as follows:

$tan(\frac{4\pi}{3})=tan(\frac{\pi}{3}+\pi)$

$=tan(\frac{\pi}{3})$

$=\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})}$

$=\frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})}$

$=\sqrt{3}$

Hence,

$tan(\frac{4\pi}{3})=\sqrt{3}$

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Example Question #2 : Sin, Cos, Tan

Calculate $cos(\frac{5\pi}{3})$ .

Possible Answers:

$-\frac{\sqrt{3}}{2}$

$-\frac{1}{2}$

$\frac{\sqrt{3}}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

First, convert the given angle measure from radians to degrees:

$cos(\frac{5\pi}{3})=cos(300^{\circ})$

Next, recall that $300^{\circ}$ lies in the fourth quadrant of the unit circle, wherein the cosine is positive. Furthermore, the reference angle of $300^{\circ}$ is

$360^{\circ}-300^{\circ}=60^{\circ}$

Hence, all that is required is to recognize from these observations that

$cos(\frac{5\pi}{3})=cos(300^{\circ}) =cos(60^{\circ})$ ,

which is $\frac{1}{2}$ .

Therefore,

$cos(\frac{5\pi}{3})=\frac{1}{2}$

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Example Question #1 : Secant, Cosecant, Cotangent

If $\cot \alpha = 5$ and $\sin \alpha < 0$ , what is the value of $\sec \alpha$ ?

Possible Answers:

$\sqrt{26}$

$-\frac{\sqrt{26}}{5}$

$\frac{\sqrt{26}}{5}$

$-\frac{1}{\sqrt{26}}$

$-\frac{5}{\sqrt{26}}$

Correct answer:

$-\frac{\sqrt{26}}{5}$

Explanation:

Since cotangent is positive and sine is negative, alpha must be in quadrant III. $\cot \alpha = \frac{x}{y}$ then implies that (x,y)=(-5,-1) is a point on the terminal side of alpha.

$r=\sqrt{(x^2+y^2)}=\sqrt{26}$

$\sec \alpha = \frac{r}{x} = \frac{\sqrt{26}}{-5}$

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Example Question #2 : Secant, Cosecant, Cotangent

If $\csc \theta < 0$ and $\tan \theta > 0$ , then which of the following must be true about $\theta$ .

Possible Answers:

$\frac{\pi}{2} < \theta < \pi$

$\frac{3\pi}{2} < \theta < 2\pi$

$\pi < \theta < \frac{3\pi}{2}$

$-\frac{\pi}{2} < \theta < 0$

$0 < \theta < \frac{\pi}{2}$

Correct answer:

$\pi < \theta < \frac{3\pi}{2}$

Explanation:

Since cosecant is negative, theta must be in quadrant III or IV.

Since tangent is positive, it must be in quadrant I or III.

Therefore, theta must be in quadrant III.

Using a unit circle we can see that quadrant III is when theta is between $\pi$ and $\frac{3\pi}{2}$ .

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Example Question #2 : Secant, Cosecant, Cotangent

The point (12, 5) lies on the terminal side of an angle in standard position. Find the secant of the angle.

Possible Answers:

$\frac{5}{12}$

$\frac{13}{5}$

$\frac{12}{13}$

$\frac{5}{13}$

$\frac{13}{12}$

Correct answer:

$\frac{13}{12}$

Explanation:

Secant is defined to be the ratio of to where is the distance from the origin.

The Pythagoreanr Triple 5, 12, 13 helps us realize that r = 13 .

Since x = 12 , the answer is $\frac{13}{12}$ .

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Example Question #3 : Secant, Cosecant, Cotangent

Given angles and in quadrant I, and given,

$\sin x = \frac{3}{5}$ and $\cos y = \frac {5}{13}$ ,

find the value of $\csc (x+y)$ .

Possible Answers:

$\frac {65}{63}$

$\frac {65}{64}$

$\frac {63}{65}$

$\frac {99}{65}$

Correct answer:

$\frac {65}{63}$

Explanation:

Use the following trigonometric identity to solve this problem.

$\csc (x+y) = \frac {1}{\sin (x+y)} = \frac {1}{\sin x \cos y + \cos x \sin y}$

Using the Pythagorean triple 3,4,5, it is easy to find $\cos x = \frac {4}{5}$ .

Using the Pythagorean triple 5,12,13, it is easy to find $\sin y = \frac {12}{13}$ .

So substituting all four values into the top equation, we get

$\csc (x+y) = \frac {1}{\frac {3}{5} \cdot \frac {5}{13} + \frac {4}{5} \cdot \frac {12}{13}} = \frac {65}{63}$

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All SAT II Math I Resources

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SAT II Math I : Trigonometry

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #3 : Understanding Sine, Cosine, And Tangent

Example Question #1 : Sine, Cosine, Tangent

Example Question #2 : Sine, Cosine, Tangent

Which of the following is equal to cos(x)?

sin(y) is equal to cos(x)

Example Question #2 : Sin, Cos, Tan

Example Question #11 : Trigonometry

Example Question #2 : Sin, Cos, Tan

Example Question #1 : Secant, Cosecant, Cotangent

Example Question #2 : Secant, Cosecant, Cotangent

Example Question #2 : Secant, Cosecant, Cotangent

Example Question #3 : Secant, Cosecant, Cotangent

All SAT II Math I Resources

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