Trigonometric Equations

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Trigonometry › Trigonometric Equations

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Which of these sine functions fulfills the following criteria?

Range of
Period of $2 \pi$
-intercept of
$f(\pi) = -1$

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

$f(x) = sin(x - \frac{\pi}{2})$

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

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

Explanation

Examining the equation of this form:

We can find , , and using the clues given:

because the range of indicates an amplitude of .
because is the midpoint between and .
because the period is not changed from the standard $2\pi$ .
$c = \frac{\pi}{2}$ . The combined fact that and $f(\pi)$ (halfway through the period) is indicates a cosine function would work - only the answer must be given as a sine function. Fortunately, though, we can use a shift-related property of $cos(x) = sin(x + \frac{\pi}{2})$ .

All told, once we realize these, then we can see that
$f(x) = sin(x + \frac{\pi}{2})$ fits our criteria.

Which of these sine functions fulfills the following criteria?

Range of
Period of $2 \pi$
-intercept of
$f(\pi) = -1$

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

$f(x) = sin(x - \frac{\pi}{2})$

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

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

Explanation

Examining the equation of this form:

We can find , , and using the clues given:

because the range of indicates an amplitude of .
because is the midpoint between and .
because the period is not changed from the standard $2\pi$ .
$c = \frac{\pi}{2}$ . The combined fact that and $f(\pi)$ (halfway through the period) is indicates a cosine function would work - only the answer must be given as a sine function. Fortunately, though, we can use a shift-related property of $cos(x) = sin(x + \frac{\pi}{2})$ .

All told, once we realize these, then we can see that
$f(x) = sin(x + \frac{\pi}{2})$ fits our criteria.

Which of these sine functions fulfills the following criteria?

Range of
Period of $2 \pi$
-intercept of
$f(\pi) = -1$

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

$f(x) = sin(x - \frac{\pi}{2})$

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

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

Explanation

Examining the equation of this form:

We can find , , and using the clues given:

because the range of indicates an amplitude of .
because is the midpoint between and .
because the period is not changed from the standard $2\pi$ .
$c = \frac{\pi}{2}$ . The combined fact that and $f(\pi)$ (halfway through the period) is indicates a cosine function would work - only the answer must be given as a sine function. Fortunately, though, we can use a shift-related property of $cos(x) = sin(x + \frac{\pi}{2})$ .

All told, once we realize these, then we can see that
$f(x) = sin(x + \frac{\pi}{2})$ fits our criteria.

A sine function where is time measured in seconds has the following properties:

Amplitude of
Minimum of
No phase shift
Frequency of Hz (cycles per second)

is which of these functions?

$f(t) = 5sin(8\pi t) + 5$

$f(t) = 5sin(8\pi t)$

$f(t) = 10sin(8\pi t) + 10$

Explanation

One important thing to realize is that the frequency is the reciprocal of the period. So if the function has a frequency of Hertz (or cycles per second), the period has to be $\frac{1}{4}$ or seconds. Because $\frac{2\pi}{b} = \frac{1}{4}$ when $b = 2\pi * 4 = 8\pi$ , we know we are looking for a equation that includes $8\pi t$ .

Also, because we have an amplitude of but a minimum of , there must be a shift upwards by units. Only one function fulfills those two criteria and the period criteria:

$f(t) = 5sin(8\pi t) + 5$

A sine function where is time measured in seconds has the following properties:

Amplitude of
Minimum of
No phase shift
Frequency of Hz (cycles per second)

is which of these functions?

$f(t) = 5sin(8\pi t) + 5$

$f(t) = 5sin(8\pi t)$

$f(t) = 10sin(8\pi t) + 10$

Explanation

Also, because we have an amplitude of but a minimum of , there must be a shift upwards by units. Only one function fulfills those two criteria and the period criteria:

$f(t) = 5sin(8\pi t) + 5$

A sine function where is time measured in seconds has the following properties:

Amplitude of
Minimum of
No phase shift
Frequency of Hz (cycles per second)

is which of these functions?

$f(t) = 5sin(8\pi t) + 5$

$f(t) = 5sin(8\pi t)$

$f(t) = 10sin(8\pi t) + 10$

Explanation

Also, because we have an amplitude of but a minimum of , there must be a shift upwards by units. Only one function fulfills those two criteria and the period criteria:

$f(t) = 5sin(8\pi t) + 5$

Factor $1-2\cos x+\cos^2 x$ .

$(1-\cos x)^2$

$\sin^2x$

$\sin^4 x$

$\cos^4 x$

$\sin x \tan x$

Explanation

Don't get scared off by the fact we're doing trig functions! Factor as you normally would. Because our middle term is negative ( $-2\cos x$ ), we know that the signs inside of our parentheses will be negative.

This means that $1-2\cos x+\cos^2 x$ can be factored to $(1-\cos x)(1-\cos x)$ or $(1-\cos x)^2$ .

Factor $1-2\cos x+\cos^2 x$ .

$(1-\cos x)^2$

$\sin^2x$

$\sin^4 x$

$\cos^4 x$

$\sin x \tan x$

Explanation

This means that $1-2\cos x+\cos^2 x$ can be factored to $(1-\cos x)(1-\cos x)$ or $(1-\cos x)^2$ .

Factor $1-2\cos x+\cos^2 x$ .

$(1-\cos x)^2$

$\sin^2x$

$\sin^4 x$

$\cos^4 x$

$\sin x \tan x$

Explanation

This means that $1-2\cos x+\cos^2 x$ can be factored to $(1-\cos x)(1-\cos x)$ or $(1-\cos x)^2$ .

Which of the following is a solution to the following equation such that $0\leq x<2\pi?$

$\frac{11\pi}{6}$

$\frac{\pi}{6}$

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

$\frac{5\pi}{4}$

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

Explanation

We begin by getting the right side of the equation to equal zero.

Next we factor.

We then set each factor equal to zero and solve.

$sin(x)=-\frac{1}{2}$

We then determine the angles that satisfy each solution within one revolution.

The angles $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ satisfy the first, and $\frac{\pi}{2}$ satisfies the second. Only $\frac{11\pi}{6}$ is among our answer choices.

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