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SAT II Math II : Graphing Trigonometric Functions

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

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Example Questions

Example Question #1 : Graphing Trigonometric Functions

Give the amplitude of the graph of the function

$f (x) = 2 \pi \sin 2 \pi x$ $\displaystyle f (x) = 2 \pi \sin 2 \pi x$

Possible Answers:

$2 \pi$ $\displaystyle 2 \pi$

$\pi$ $\displaystyle \pi$

$\displaystyle 2$

$4 \pi$ $\displaystyle 4 \pi$

$\displaystyle 1$

Correct answer:

$2 \pi$ $\displaystyle 2 \pi$

Explanation:

The amplitude of the graph of a sine function $f(x) = A \sin Bx$ $\displaystyle f(x) = A \sin Bx$ is $\displaystyle A$. Here, $A= 2 \pi$ $\displaystyle A= 2 \pi$, so this is the amplitude.

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Example Question #2 : Graphing Trigonometric Functions

Which of these functions has a graph with amplitude 4?

Possible Answers:

$f(x) = 4 \cos \frac{\pi x}{5}$ $\displaystyle f(x) = 4 \cos \frac{\pi x}{5}$

$f(x) = \frac{1}{7}\cos \frac{\pi x }{4}$ $\displaystyle f(x) = \frac{1}{7}\cos \frac{\pi x }{4}$

$f(x) = \frac{1}{4}\cos \frac{\pi x}{3}$ $\displaystyle f(x) = \frac{1}{4}\cos \frac{\pi x}{3}$

$f(x) = 9\cos 4x$ $\displaystyle f(x) = 9\cos 4x$

$f(x) = \frac{2}{3}\cos \frac{ x }{4}$ $\displaystyle f(x) = \frac{2}{3}\cos \frac{ x }{4}$

Correct answer:

$f(x) = 4 \cos \frac{\pi x}{5}$ $\displaystyle f(x) = 4 \cos \frac{\pi x}{5}$

Explanation:

The functions in each of the choices take the form of a cosine function

$f(x)= A \cos Nx$ $\displaystyle f(x)= A \cos Nx$.

The graph of a cosine function in this form has amplitude $\displaystyle A$. Therefore, for this function to have amplitude 4, A = 4 $\displaystyle A = 4$. Of the five choices, only

$f(x) = 4 \cos \frac{\pi x}{5}$ $\displaystyle f(x) = 4 \cos \frac{\pi x}{5}$

matches this description.

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Example Question #1 : Graphing Trigonometric Functions

Which of these functions has a graph with amplitude $\frac{4}{7}$ $\displaystyle \frac{4}{7}$ ?

Possible Answers:

$f(x)=\sin \frac{7x}{4}$ $\displaystyle f(x)=\sin \frac{7x}{4}$

$f(x)=\sin \frac{4x}{7}$ $\displaystyle f(x)=\sin \frac{4x}{7}$

$f(x)= \frac{4}{7} \sin x$ $\displaystyle f(x)= \frac{4}{7} \sin x$

$f(x)=\sin \frac{4 \pi x}{7}$ $\displaystyle f(x)=\sin \frac{4 \pi x}{7}$

$f(x)= \frac{2}{7} \sin x$ $\displaystyle f(x)= \frac{2}{7} \sin x$

Correct answer:

$f(x)= \frac{4}{7} \sin x$ $\displaystyle f(x)= \frac{4}{7} \sin x$

Explanation:

The functions in each of the choices take the form of a sine function

$f(x)= A \sin Bx$ $\displaystyle f(x)= A \sin Bx$.

The graph of a sine function in this form has amplitude $\displaystyle A$. Therefore, for this function to have amplitude 4, $A = \frac{4}{7}$ $\displaystyle A = \frac{4}{7}$. Of the five choices, only

$f(x)= \frac{4}{7} \sin x$ $\displaystyle f(x)= \frac{4}{7} \sin x$

matches this description.

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Example Question #1 : Graphing Trigonometric Functions

Which of the following sine functions has a graph with period of 7?

Possible Answers:

$f(x)= \frac{11}{5} \sin \frac{7 x}{2}$ $\displaystyle f(x)= \frac{11}{5} \sin \frac{7 x}{2}$

$f(x)= \frac{11}{5} \sin \frac{2 x}{7}$ $\displaystyle f(x)= \frac{11}{5} \sin \frac{2 x}{7}$

$f(x)= \frac{11}{5} \sin \frac{ \pi x}{7}$ $\displaystyle f(x)= \frac{11}{5} \sin \frac{ \pi x}{7}$

$f(x)= \frac{11}{5} \sin \frac{7 \pi x}{2}$ $\displaystyle f(x)= \frac{11}{5} \sin \frac{7 \pi x}{2}$

$f(x)= \frac{11}{5} \sin \frac{2 \pi x}{7}$ $\displaystyle f(x)= \frac{11}{5} \sin \frac{2 \pi x}{7}$

Correct answer:

$f(x)= \frac{11}{5} \sin \frac{2 \pi x}{7}$ $\displaystyle f(x)= \frac{11}{5} \sin \frac{2 \pi x}{7}$

Explanation:

The period of the graph of a sine function $f(x)= A \sin B x$ $\displaystyle f(x)= A \sin B x$, is $\frac{2 \pi}{B}$ $\displaystyle \frac{2 \pi}{B}$, or $2 \pi \div B$ $\displaystyle 2 \pi \div B$.

Therefore, we solve for $\displaystyle B$:

$\frac{2 \pi}{B} = 7$ $\displaystyle \frac{2 \pi}{B} = 7$

$\frac{B}{2 \pi} = \frac{1}{7}$ $\displaystyle \frac{B}{2 \pi} = \frac{1}{7}$

$B= \frac{1}{7} \cdot 2 \pi = \frac{2 \pi}{7}$ $\displaystyle B= \frac{1}{7} \cdot 2 \pi = \frac{2 \pi}{7}$

The correct choice is therefore $f(x)= \frac{11}{5} \sin \frac{2 \pi x}{7}$ $\displaystyle f(x)= \frac{11}{5} \sin \frac{2 \pi x}{7}$.

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Example Question #1 : Period And Amplitude

Which of the given functions has the greatest amplitude?

Possible Answers:

$\sin(4x)$ $\displaystyle \sin(4x)$

$\sin(x+3)$ $\displaystyle \sin(x+3)$

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

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

$2\sin(x)$ $\displaystyle 2\sin(x)$

Correct answer:

$2\sin(x)$ $\displaystyle 2\sin(x)$

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 $2\sin(x)$ $\displaystyle 2\sin(x)$.

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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SAT II Math II : Graphing Trigonometric Functions

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

All SAT II Math II Resources

Example Questions

Example Question #1 : Graphing Trigonometric Functions

Example Question #2 : Graphing Trigonometric Functions

Example Question #1 : Graphing Trigonometric Functions

Example Question #1 : Graphing Trigonometric Functions

Example Question #1 : Period And Amplitude

All SAT II Math II Resources

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