### All Trigonometry Resources

## Example Questions

### Example Question #1 : Understanding Trigonometric Equations

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

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

**Possible Answers:**

**Correct answer:**

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.

### Example Question #1 : Understanding Trigonometric Equations

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

**Possible Answers:**

**Correct answer:**

Rearrange algebraically so that,

.

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

### Example Question #1 : Setting Up Trigonometric Equations

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

**Possible Answers:**

**Correct answer:**

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

.

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.

### Example Question #4 : Trigonometric Equations

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

**Possible Answers:**

**Correct answer:**

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.

### Example Question #1 : Setting Up Trigonometric Equations

Solve each equation over the interval

**Possible Answers:**

**Correct answer:**

Rearrange the equation so that,

.

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

.

These measures over the interval

.

### Example Question #161 : Trigonometry

Solve each quation over the interval

**Possible Answers:**

**Correct answer:**

Rearrange the equation so that,

.

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

.

These two angles are and .

### Example Question #7 : Trigonometric Equations

Solve for using trigonometric ratios.

**Possible Answers:**

**Correct answer:**

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:

The sine of is , so we can substitue that in:

cross multiplying gives us .

### Example Question #8 : Trigonometric Equations

Solve for using trigonometric ratios.

**Possible Answers:**

**Correct answer:**

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:

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 is , and the cosine is . Find the tangent by dividing:

Now we can substitute that value into the original equation we set up:

multiply both sides by 3

### Example Question #9 : Trigonometric Equations

Solve for using trigonometric ratios.

**Possible Answers:**

**Correct answer:**

Start by setting up the trigonometric ratio using the angles and sides given. We have the side length adjacent to the angle, 1, and the hypotenuse, x, so we will be using cosine:

The cosine of is , so this makes our equation now:

. x must equal 2.

### Example Question #1 : Setting Up Trigonometric Equations

Which of these sine functions fulfills the following criteria?

- Range of
- Period of
- -intercept of

**Possible Answers:**

**Correct answer:**

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 .
- . The combined fact that and (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 .

All told, once we realize these, then we can see that

fits our criteria.

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