Trigonometry - Setting Up Trigonometric Equations

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 triangle has sides , , of lengths , , respectively. The angle opposite each side is called , , , respectively. The sine of which angle and the cosine and which different angle will both yield $\frac{8}{17}$

In the answer, list the sine first and the cosine second.

Explanation

Screen_shot_2015-03-07_at_2.36.12_pm

This is the figure being described in the problem, and as sine is opposite over hypotenuse and cosine adjacent over hypotenuse, the sine of and the cosine of will yield the correct answer.

$\sin(a)=\frac{opp}{hyp}=\frac{8}{17}$

$\cos(b)=\frac{adj}{hyp}=\frac{8}{17}$

Solve the equation over the domain (answer in degrees).

Explanation

Rearrange algebraically so that,

Over the interval 0 to 360 degrees, sinx = 1 when x equals 90 degrees.

Solve each equation over the domain (answer in degrees).

Explanation

First, think of the angle values for which . (This is the equivalent of taking the arctan.)

$x-15=tan^{-1}(1)$

$x=tan^{-1}+15$ .

The angles for which this is true are 45 degrees and 225 degrees.

We set x-15 equal to those two angles and solve for x, giving us 60 and 240.

Solve each equation over the domain (answer in degrees).

Explanation

Rearrange the equation so that,

Recall the angles over the interval 0 to 360 degrees for which sec is equal to 2.

These are 60 and 300 degrees.

Set x+7 equal to these angle measures and then find that x equals 53 and 293.

Solve each equation over the interval $2\pi$

$\frac{\pi}{4}, \frac{7\pi}{4}$

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

$\frac{3\pi}{4}, \frac{5\pi}{4}$

$\frac{\pi}{6}, \frac{7\pi}{6}$

Explanation

Rearrange the equation so that,

$cos^2x = \frac{1}{2}$ .

Take the square of both sides and then recall the angle measures for which,

$cos(x)={\frac{1}{\sqrt2}}$ .

These measures over the interval $0<x< 2 \pi$

$x=\frac{\pi}{4}; \frac{7\pi}{4}$ .

Solve each quation over the interval $0 < x < 2\pi$

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

$0, : 2\pi$

$\frac{\pi}{4}, \frac{7\pi}{4}$

$0, \frac{\pi}{2}$

Explanation

Rearrange the equation so that,

Take the square of both sides and find the angles for which

$sin(x)=\pm1$ .

These two angles are $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ .

Solve for using trigonometric ratios.

Set up trig 1

$6\sqrt2$

$\frac{1}{\sqrt2}$

$\sqrt{12}$

Explanation

To solve for x, first set up a trigonometric equation using the information provided in the diagram. The two side lengths given are the hypotenuse, x, and the side opposite the given angle, 6. We can set up our equation like this:

$sin\left(\frac{\pi}{4}\right)= \frac{6}{x}$

The sine of $\frac{\pi}{4}$ is $\frac{1}{\sqrt2}$ , so we can substitue that in:

$\frac{1}{\sqrt2}=\frac{6}{x}$

cross multiplying gives us $x=6\sqrt2$ .

Solve for using trigonometric ratios.

Set up trig 2

$3\sqrt3$

$\frac{3\sqrt3}{2}$

$\frac{3}{2}$

$\frac{3}{\sqrt3}$

Explanation

To solve for x, first set up a trigonometric equation using the information provided in the diagram. The two side lengths given are x, the side opposite the angle, and 3, the side adjacent to the angle. This means we'll be using tangent. Set up the equation like this:

$tan\left(\frac{\pi}{3}\right)=\frac{x}{3}$

We can't just know the tangent by using the unit circle, but we can easily figure it out using sine and cosine. Tangent can be evaluated as sine over cosine.

The sine of $\frac{\pi}{3}$ is $\frac{\sqrt3}{2}$ , and the cosine is $\frac{1}{2}$ . Find the tangent by dividing: