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High School Math : Trigonometry

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Example Questions

Example Question #3 : Trigonometric Functions And Graphs

What is the amplitude of $y=3\sin(2x+5)$ ?

Possible Answers:

$-\frac{5}{2}$

$-\frac{5}{3}$

Correct answer:

Explanation:

The amplitude of a wave function like $y=3\sin(2x+5)$ is always going to be the coefficient of the function. In this case, that is .

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Example Question #1 : Understanding Period And Amplitude

What is the local maximum of $\sin(2x)$ between $-\pi$ and $\pi$ ?

Possible Answers:

0.05

Correct answer:

Explanation:

The fastest way to solve this problem is to graph it and observe the answer. However, the other option is to think of this equation in terms of period.

When the coefficient of the variable increases, the frequency increases and the period decreases by that rate.

Since our equation is $\sin(2x)$ , our period will be $\frac{1}{2}$ the normal period of a $\sin(x)$ wave. Since only the period is changing, the amplitude is not. Therefore the amplitude (the highest and lowest points) of $\sin(2x)$ will be the same as that of $\sin(x)$ . The amplitude of a sine wave is , so the amplitude of $\sin(2x)$ will also be .

Therefore, our maximum will be .

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Example Question #3 : Period And Amplitude

Which of the given functions has the greatest amplitude?

Possible Answers:

$\sin(4x)$

$\frac{1}{2}\sin(4x)$

$\sin(x+3)$

$\sin\frac{2\pi }{3}(x+2)$

$2\sin(x)$

Correct answer:

$2\sin(x)$

Explanation:

The amplitude of a function is the amount by which the graph of the function travels above and below its midline. When graphing a sine function, the value of the amplitude is equivalent to the value of the coefficient of the sine. Similarly, the coefficient associated with the x-value is related to the function's period. The largest coefficient associated with the sine in the provided functions is 2; therefore the correct answer is $2\sin(x)$ .

The amplitude is dictated by the coefficient of the trigonometric function. In this case, all of the other functions have a coefficient of one or one-half.

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Example Question #1 : Trigonometric Identities

Trig_id

What is the $\tan$ of $\theta$ ?

Possible Answers:

$\frac{o}{a}$

$\frac{a}{h}$

$\frac{o}{h}$

$\frac{a}{o}$

$\frac{a}{h}$

Correct answer:

$\frac{o}{a}$

Explanation:

When working with basic trigonometric identities, it's easiest to remember the mnemonic: SOHCAHTOA .

$\sin=\frac{opposite}{hypotenuse}$

$\cos=\frac{adjacent}{hypotenuse}$

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

When one names the right triangle, the opposite side is opposite to the angle, the adjacent side is next to the angle, and the hypotenuse spans the two legs of the right angle.

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Example Question #2 : Trigonometric Identities

Simplify $(1-\cos^2\theta )\cdot\cot\theta\cdot\csc^2\theta$ .

Possible Answers:

$\cos^2\theta$

$\sin\theta$

$\sin^2\theta$

$\cot\theta$

$\tan\theta$

Correct answer:

$\cot\theta$

Explanation:

Simplifying trionometric expressions or identities often involves a little trial and error, so it's hard to come up with a strategy that works every time. A lot of times you have to try multiple strategies and see which one helps.

Often, if you have any form of $\tan\theta,$ $\cot\theta,$ $\csc\theta,$ or $\sec\theta,$ in an expression, it helps to rewrite it in terms of sine and cosine. In this problem, we can use the identities $\cot\theta=\frac{\cos\theta }{\sin\theta }$ and $\csc\theta=\frac{1}{\sin\theta }$ .

$(1-\cos^2\theta )\cdot\cot\theta\cdot\csc^2\theta=(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot(\frac{1}{\sin\theta})^2$

$=(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}$ .

This doesn't seem to help a whole lot. However, we should recognize that $(1-\cos^2\theta )=\sin^2\theta$ because of the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ .

$(1-\cos^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}=(\sin^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}$

We can cancel the $\sin^2\theta$ terms in the numerator and denominator.

$(\sin^2\theta )\cdot\frac{\cos\theta }{\sin\theta }\cdot\frac{1}{\sin^2\theta}=\frac{\cos\theta }{\sin\theta }=\cot\theta$ .

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Example Question #2 : Trigonometric Identities

Trig_id

What is the $\sin$ of $\theta$ ?

Possible Answers:

$\frac{h}{a}$

$\frac{o}{a}$

$\frac{a}{h}$

$\frac{o}{h}$

$\frac{h}{o}$

Correct answer:

$\frac{o}{h}$

Explanation:

When working with basic trigonometric identities, it's easiest to remember the mnemonic: SOHCAHTOA .

$\sin=\frac{opposite}{hypotenuse}$

$\cos=\frac{adjacent}{hypotenuse}$ ,

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

When one names the right triangle, the opposite side is opposite to the angle, the adjacent side is next to the angle, and the hypotenuse spans the two legs of the right angle.

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Example Question #3 : Trigonometric Identities

Trig_id

What is the $\cos$ of $\theta$ ?

Possible Answers:

$\frac{h}{o}$

$\frac{h}{a}$

$\frac{a}{h}$

$\frac{o}{h}$

$\frac{o}{a}$

Correct answer:

$\frac{a}{h}$

Explanation:

When working with basic trigonometric identities, it's easiest to remember the mnemonic: SOHCAHTOA .

$\sin=\frac{opposite}{hypotenuse}$

$\cos=\frac{adjacent}{hypotenuse}$ ,

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

When one names the right triangle, the opposite side is opposite to the angle, the adjacent side is next to the angle, and the hypotenuse spans the two legs of the right angle.

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Example Question #1 : Trigonometric Identities

Simplify

$\cot(x)\times{\sin(x) }$

Possible Answers:

$\cos(x)$

$\csc(x)$

$\arcsin(\pi )$

$\arcsin(x)$

$\tan(x)$

Correct answer:

$\cos(x)$

Explanation:

$\cot(x) = \cos(x)/\sin(x)$ . Thus: $[\cos(x)/\sin(x)]\times\sin(x)=\cos(x)$

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Example Question #1 : Using Pythagorean Identities

Simplify

$\frac{cot (x)\cdot tan(x)}{cos(x)}$

Possible Answers:

$\cos(x)$

$\cot(x)/sec(x)$

$\sec(x)$

$\csc(x)$

$\cos(x)\times\tan(x)$

Correct answer:

$\sec(x)$

Explanation:

$\cot(x)\times\tan(x) = 1$

and

$1/(\cos(x)) = \sec(x)$ .

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Example Question #6 : Trigonometric Identities

Simplify $\frac{\cos^2x-1}{\sin x}$ .

Possible Answers:

$-\sin x$

$\sin^2 x$

$\sin x$

$\sin x\cos x$

$\tan x$

Correct answer:

$-\sin x$

Explanation:

Remember that $\sin^2x+\cos^2 x=1$ . We can rearrange this to simplify our given equation:

$\frac{\cos^2x-1}{\sin x}$

$=\frac{-\sin^2x}{\sin x}=-\sin x$

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High School Math : Trigonometry

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #3 : Trigonometric Functions And Graphs

Example Question #1 : Understanding Period And Amplitude

Example Question #3 : Period And Amplitude

Example Question #1 : Trigonometric Identities

Example Question #2 : Trigonometric Identities

Example Question #2 : Trigonometric Identities

Example Question #3 : Trigonometric Identities

Example Question #1 : Trigonometric Identities

Example Question #1 : Using Pythagorean Identities

Example Question #6 : Trigonometric Identities

All High School Math Resources

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