# Evaluating Trig Functions – Explanation and Examples

*Evaluating trig functions exactly and without a calculator involves memorizing a few trigonometric values and identities.*

Knowing just the trig ratios in the first quadrant is enough to find angles in other quadrants with just a few tools. While calculators can give decimal answers, these methods give exact results.

As with all trigonometric topics, evaluating trig functions is important for physical sciences, architecture, and engineering.

Before we continue with our article, you should make sure you have completely memorized the unit circle and understood trig functions.

This section covers:

**How to Evaluate Trig Functions**

**Important Trig Values****Important Trig Identities**

**How to Evaluate Trigonometric Functions**

Evaluating trig functions without a calculator is possible. Doing so requires a bit more work, but it results in more exact answers.

Evaluating trig functions uses all of the skills learned previously, including:

- Memorizing angles on the unit circle
- Memorizing sine and cosine values for quadrantal angles and major first quadrant angles
- Recalling the sign of sine and cosine in each quadrant
- Memorizing trigonometric identities, especially double angle, half-angle, sum, and difference identities.

## Important Trig Values

The most important trig values are the sine and cosine of the major angles in the first quadrant and the quadrantal angles.

These are:

Angle | 0 | $\frac{\pi}{6}$ | $\frac{\pi}{4}$ | $\frac{\pi}{3}$ | $\frac{\pi}{2}$ | $\pi$ | $\frac{3\pi}{2}$ |

Sine | 0 | $\frac{1}{\sqrt{2}}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | 1 | 0 | -1 |

Cosine | 1 | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{\sqrt{2}}$ | 0 | -1 | 0 |

Since tangent is equal to sine divided by cosine, these values are enough to determine tangent. Then, the other three trig functions are just reciprocals of the first three.

After that, knowing that the absolute value of sine and cosine are the same at corresponding angles in the other quadrants is important. Only the sign will change for angles that are the same distance from the x-axis.

The mnemonic All Students Take Calculus helps with this. Here, A stands for All, S stands for Sine, T stands for Tangent, and C stands for Cosine. This smart phrase expresses which of the three main trig functions (sine, cosine, and tangent) are positive in which quadrant. All are positive in the first, then only sine in the second, then only tangent in the third, and finally only cosine in the fourth.

## Important Trig Identities

After this, finding minor angles requires using trig identities. The most useful are the double angle, half-angle, sum, and difference identities for sine and cosine.

Most of these are a little complex, so knowing the identities $sin^2+cos^2 = 1$ is a shortcut for finding cosine given sine or sine given cosine.

The double angle identities are:

- $sin(2x) = 2sinxcosx = \frac{2tanx}{1+tan^2x}$.
- $cos(2x) = cos^2x-sin^2x = 2cos^2x-1 = 1-2sin^2x = \frac{1-tan^2x}{1+tan^2x}$.

There is more than one double angle formula for both sine and cosine, so use whichever one has the easiest calculations.

The half-angle identities are:

- $sin(\frac{x}{2}) = \sqrt{\frac{1-cosx}{2}}$
- $cos(\frac{x}{2}) = \sqrt{\frac{1+cosx}{2}}$.

Then, the sum and difference identities are:

- $sin(\theta_1+\theta_2) = sin(\theta_1)cos(\theta_2) + sin(\theta_2)cos(\theta_1)$.
- $cos(\theta_1+\theta_2) = cos(\theta_1)cos(\theta_2) – sin(\theta_2)sin(\theta_1)$.

To turn these into the difference identities, change addition signs to subtraction signs and, for cosine, change the subtraction sign on the right side to a plus sign.

## Examples

This section goes over common examples of problems involving evaluating trig functions and their step-by-step solutions.

### Example 1

Find the sine of the angle $\frac{7\pi}{6}$ radians.

### Solution

First, note that this angle is in the third quadrant. Since tangent is the only one of the main three trigonometric ratios that is positive in the third quadrant, sine is negative there.

Now, this angle is $\frac{\pi}{6}$ units from the x-axis. This means that the sine value of this angle will have the same absolute value as the sine of $\frac{\pi}{6}$ in the first quadrant. By the left-hand trick, this sine is $\frac{1}{2}$.

Therefore, the sine of the angle $\frac{7\pi}{6}$ radians is $-\frac{1}{2}$.

### Example 2

Find the cosine of $-\frac{3\pi}{2}$.

### Solution

Finding the cosine of this angle involves finding the cosine of the equivalent standard angle.

To find this, subtract $\frac{3\pi}{2}$ from a full angle, $2\pi$ radians. This is $\frac{\pi}{2}$ radians.

Therefore, the cosine of this angle is equal to the cosine of $\frac{\pi}{2}$ radians, which is $0$.

### Example 3

Find secant of $\frac{\pi}{8}$ radians using a half-angle identity.

### Solution

Since secant is equal to the reciprocal of the cosine, it is required to first find the cosine of $\frac{\pi}{8}$ radians.

Note that $\frac{\pi}{8} = \frac{1}{2}\frac{\pi}{4}$. Therefore, use the half-angle identity for cosine, using $\frac{\pi}{4}$ as the original angle.

Now, recall that the half-angle identity for cosine is:

$cos(\frac{x}{2}) = \sqrt{\frac{1+cosx}{2}}$.

Therefore, in this case, the cosine of $\frac{\pi}{8}$ is:

$\sqrt{1+cos(\frac{\pi}{4})}{2}} = \sqrt{1+\frac{\sqrt{2}}{2}{2}} = \sqrt{\frac{\sqrt{2}+2}{2}\frac{1}{2}} = \sqrt{\frac{\sqrt{2}+2}{4}} = \frac{1}{2}\sqrt{\sqrt{2}+2}.

### Example 4

Find the cosecant of the angle $\frac{25\pi}{12}$ radians.

### Solution

Recall that the cosecant is the reciprocal of the sine. Therefore, to find this ratio, begin by finding the sine.

To find this ratio, it is important to notice that the angle $\frac{25\pi}{12}$ radians is equal to $\frac{7\pi}{4}+\frac{\pi}{3}$. The sine of both of these angles is known, so it is possible to use the angle sum formula here.

Let $\frac{7\pi}{4}$ be the first angle and let $frac{\pi}{3}$ be the second. Then, the sine of the angle $\frac{25\pi}{12}$ radians is:

$sin(\frac{7\pi}{4})cos(\frac{\pi}{3}) + sin(\frac{\pi}{3})cos(\frac{7\pi}{4})$.

Plugging in the function values, this is:

$-\frac{\sqrt{2}}{2}\frac{1}{2} + \frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$,

This simplifies to:

$-\frac{\sqrt{2}}{4}+\frac{\sqrt{3}\sqrt{2}}{4} = \frac{\sqrt{2}(\sqrt{3}-1)}{4}$.

Then, the cosecant is the reciprocal of this:

$\frac{4}{sqrt{2}(\sqrt{3}-1)}$.

### Example 5

Describe how to find the tangent of the angle $\frac{5\pi}{48}$.

### Solution

There are several ways to do this. This way will use the given formulas for sine and cosine, but it is possible to use them to derive a simple formula for the tangent, which would be the most straightforward way to do this problem.

First, note that $\frac{5\pi}{48} = \frac{2\pi}{48}+\frac{3\pi}{48} = \frac{\pi}{24}+\frac{\pi}{16}$.

But, $\frac{\pi}{24} = \frac{1}{2}\frac{1}{2}\frac{\pi}{6}$ and $\frac{\pi}{16} = \frac{1}{2}\frac{1}{2}\frac{\pi}{4}$.

Therefore, using the half-angle formula for sine and cosine twice each for $\frac{\pi}{6}$ radians and then again twice each for $\frac{\pi}{4}$ will produce four ratios.

Then, use the sine and cosine of $\frac{\pi}{24}$ radians and $\frac{\pi}{16}$ radians in the sum formula for sine. Do this again with cosine.

Now, divide the sine value by the cosine value. This will produce the tangent.

### Practice Questions

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