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New SAT Math - Calculator : Sine, Cosine, & Tangent

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Example Questions

Example Question #1 : Trigonometry

Find the value of the trigonometric function in fraction form for triangle ABC .

Triangle

What is the cosine of $\angle B$ ?

Possible Answers:

$\frac{7}{25}$

24/25

7/24

Correct answer:

$\frac{7}{25}$

Explanation:

The cosine of an angle is the value of the adjacent side over the hypotenuse.

Therefore:

$cos \angle B = \frac{adjacent}{hypotenuse} = \frac{7}{25}$

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Example Question #2 : Trigonometry

If cos x = 0.2 $\small \cosx$ and sin x = 0.4, what is the value of tan x?

Possible Answers:

0.035

Correct answer:

Explanation:

$\small \tan{x}=\frac{\sin{x}}{\cos{x}}$

$\small \small \tan{x}=\frac{0.4}{0.2}$

$\small \tan{x}=0.035$

$\small x=\tan^{-1}0.035$

$\small x=2$

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Example Question #2 : Sin, Cos, Tan

What is the value of sin(30)+sin(60) ?

Possible Answers:

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

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

Correct answer:

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

Explanation:

Solve each term separately.

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

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

Add both terms.

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

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Example Question #2 : Sin, Cos, Tan

Determine the value of 2tan(120) .

Possible Answers:

$-\sqrt3$

$-2\sqrt3$

$2\sqrt3$

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

$\sqrt3$

Correct answer:

$-2\sqrt3$

Explanation:

Rewrite 2tan(120) in terms of sines and cosines.

$2tan(120)=2(\frac{sin(120)}{cos(120)})=2(\frac{\frac{\sqrt3}{2}}{-\frac{1}{2}})$

Simplify the complex fraction.

$2(\frac{\frac{\sqrt3}{2}}{-\frac{1}{2}})= 2\times\frac{\sqrt3}{2}\times-2 = -2\sqrt3$

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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 #1 : Sin, Cos, Tan

Select the ratio that would give Tan B.

Possible Answers:

$\tan B=\frac{CB}{AB}$

$\tan B=\frac{AC}{AB}$

None of the other answers.

$\tan B=\frac{AB}{AC}$

$\tan B=\frac{CB}{AC}$

Correct answer:

$\tan B=\frac{AC}{AB}$

Explanation:

We need the Tan B. Which side lengths correspond to this ratio?

$\tan=\frac{opp}{adj}$

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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 #2 : Sin, Cos, Tan

What is the result when the following expression is simplified as much as possible?

$\sin(2h)\sec(h)+2\sin(-h)$

Possible Answers:

$2\sin h$

$\sec h$

Correct answer:

Explanation:

Because $\sin x$ is an odd function, we can rewrite the second term in the expression.

$2\sin(-h)=-2\sin h$ .

We now use a double-angle formula to expand the first term.

$\sin(2h)\sec h=2\sin h\cos h\sec h$ .

Because they are reciprocals, $\cos h\sec h=1$ .

$2\sin h\cos h\sec h-2\sin h=2\sin h-2\sin h=0.$

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Example Question #1 : Sine, Cosine, & Tangent

Round to the nearest hundredth.

Use your calculator to find:

$\sin72^{\circ}$

Possible Answers:

None of the above

0.95

0.72

0.25

Correct answer:

0.95

Explanation:

Before plugging the function into the calculator make sure the mode of the calculator is set to degrees,

Plug in $\sin72^{\circ}$ which equals to 0.95 .

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New SAT Math - Calculator : Sine, Cosine, & Tangent

Study concepts, example questions & explanations for New SAT Math - Calculator

All New SAT Math - Calculator Resources

Example Questions

Example Question #1 : Trigonometry

Example Question #2 : Trigonometry

Example Question #2 : Sin, Cos, Tan

Example Question #2 : Sin, Cos, Tan

Example Question #2 : Sin, Cos, Tan

Example Question #1 : Sin, Cos, Tan

Example Question #11 : Trigonometry

Example Question #2 : Sin, Cos, Tan

Example Question #2 : Sin, Cos, Tan

Example Question #1 : Sine, Cosine, & Tangent

All New SAT Math - Calculator Resources

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