Sine, Cosine, Tangent

SAT Subject Test in Math II · Learn by Concept

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SAT Subject Test in Math II › Sine, Cosine, Tangent

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1

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

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

CORRECT

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

0

0

0

0

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

2

Find the value of in exact form.

$5\sqrt{3}$

CORRECT

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

0

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

0

$10\sqrt3$

0

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

0

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: $5\sqrt{3}$

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