Trigonometry

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SAT Math › Trigonometry

Questions 1 - 10

1

Determine the exact value of .

$4\sqrt2$

$8\sqrt2$

$6\sqrt2$

$\frac{\sqrt2}{8}$

$\frac{\sqrt2}{4}$

Explanation

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

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

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

2

Solve for $\theta$ between $[0,2\pi]$ .

$\cos (2\theta )=\frac{1}{2}$

$\frac{\pi}{6},\frac{5\pi}{6}$

$\frac{\pi}{3},\frac{5\pi}{3}$

$\frac{\pi}{3},\frac{2\pi}{3}$

$\frac{\pi}{12},\frac{5\pi}{12}$

Explanation

First we must solve for when sin is equal to 1/2. That is at

$\frac{\pi}{3},\frac{5\pi}{3}$

Now, plug it in:

$2\theta =\frac{\pi}{3}\Rightarrow \theta =\frac{\pi}{6},$

$2\theta =\frac{5\pi}{3}\Rightarrow \theta =\frac{5\pi}{6}$

3

In a triangle, , what is the measure of angle A if the side opposite of $\angleA$ angle A is 3 and the adjacent side to angle A is 4?

(Round answer to the nearest tenth of a degree.)

$36.9^{\circ}$

$40^{\circ}$

$32^{\circ}$

$34.7^{\circ}$

Explanation

To find the measure of angle of A we will use tangent to solve for A. We know that

$tan A=\frac{opposite}{adjacent}$

In our case opposite = 3 and adjacent = 4, we substitute these values in and get:

$tan A= \frac{3}{4}$

Now we take the inverse tangent of each side to find the degree value of A.

$A= tan^{^{-1}}\left(\frac{3}{4}\right)$

4

Sine

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

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)

5

If , what is if is between and $2\pi$ ?

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

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

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

6

What is the value of ?

$\frac{\sqrt3+1}{2}$

$\frac{\sqrt3-1}{2}$

Explanation

Solve each term separately.

$cos(30)= \frac{\sqrt3}{2}$

$cos(60)=\frac{1}{2}$

Add both terms.

$\frac{\sqrt3}{2}+\frac{1}{2}= \frac{\sqrt3+1}{2}$

7

If , what is if is between and $2\pi$ ?

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

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

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

8

What is the value of ?

$\frac{\sqrt3+1}{2}$

$\frac{\sqrt3-1}{2}$

Explanation

Solve each term separately.

$cos(30)= \frac{\sqrt3}{2}$

$cos(60)=\frac{1}{2}$

Add both terms.

$\frac{\sqrt3}{2}+\frac{1}{2}= \frac{\sqrt3+1}{2}$

9

Find the value of in exact form.

$5\sqrt{3}$

$\frac{5\sqrt{3}}{2}$

$\frac{5\sqrt{3}}{3}$

$10\sqrt3$

$\frac{10\sqrt3}{3}$

Explanation

Recall that:

$tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$

This means that: $tan(60) = \frac{sin(60)}{cos(60)}$

$sin(60) = \frac{\sqrt3}{2}$

$cos(60) = \frac{1}{2}$

Divide the two terms.

$\frac{\sqrt3}{2} \div \frac{1}{2} = \frac{\sqrt3}{2} \times \frac{2}{1} = \sqrt3$

This means that $5tan(60) = 5\sqrt{3}$ .

The answer is: $5\sqrt{3}$

10

Solve for $\theta$ between $[0,2\pi]$ .

$\sin (2\theta )=\frac{1}{2}$

$\frac{\pi}{12},\frac{5\pi}{12}$

$\frac{\pi}{6},\frac{5\pi}{6}$

$\frac{\pi}{3},\frac{2\pi}{3}$

$\frac{\pi}{6},\frac{\pi}{3}$

Explanation

First we must solve for when sin is equal to 1/2. That is at

$\frac{\pi}{6},\frac{5\pi}{6}$

Now, plug it in:

$2\theta =\frac{\pi}{6}\Rightarrow \theta =\frac{\pi}{12}$

$2\theta =\frac{5\pi}{6}\Rightarrow \theta =\frac{5\pi}{12}$

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