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Learning Objectives

- Find function values for the sine and cosine of \(30^{\text{o}}\) or \(\dfrac{\pi}{6}\), \(45^{\text{o}}\) or \(\dfrac{\pi}{4}\), and \(60^{\text{o}}\) or \(\dfrac{\pi}{3}\).
- Find reference angles.
- Use reference angles to evaluate trigonometric functions.

##### Be Prepared

Before you get started, take this readiness quiz.

- Draw a unit circle:\(x^2+y^2=1\).
- Determine how many points on the unit circle have \(-\dfrac13\) as their \(x\)-coordinate. Indicate these on the graph.
- Determin how many points on the unit circle have \(-\dfrac13\) as their \(y\)-coordinate. Indicate these on the graph.

In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.

## Defining Sine, Cosine, and Tangent ratios for any angle

To define our trigonometric ratios, we begin by drawing a unit circle (a circle of radius \(1\) centered at the origin \((0,0)\)).

Recall that the *x- *and *y-*axes divide the coordinate plane into four quarters called quadrants. We label these quadrants to mimic the direction a positive angle would sweep. The four quadrants are labeled I, II, III, and IV.

For an acute angle \(t,\) we can label the intersection of the terminal side and the unit circle as by its coordinates, \((x,y)\). Since the circle has radius \(1\), using our trigonometric ratios we see that the triangle in orange below has base \(\cos t\) and height \(\sin t\). The coordinates \(x\) and \(y\) therefore can be related to the angle \(t\):\(x= \cos t\) and \(y= \sin t\). Also, \(\tan t=\dfrac{\sin t}{\cos t}\).

UNIT CIRCLE

A unit circle has a center at \((0,0)\) and radius \(1\). Form the angle with measure \(t\) with initial side coincident with the \(x\)-axis.

Let \((x,y)\) bepoint where the terminal side of the angle and unit circle meet. Then \((x,y)=(\cos t,\sin t)\). Further, \(\tan t=\dfrac{\sin t}{\cos t}\).

### Defining Sine, Cosine, and Tangent for any Angle

Now that we have our unit circle labeled, we can learn how the \((x,y)\) coordinates relate to the angle \(t\). For an acute angle, the **sine ratio **\(\sin t\) isthe \(y\)-coordinate of the point where the corresponding terminal side of the angle intersects the unit circle and the **cosine ratio **\(\cost\) isthe \(x\)-coordinate of the point where the corresponding terminal side of the angle intersects the unit circle.We extend this definition to all angles. So, we definethe **sine ratio **\(\sin t\)to bethe \(y\)-coordinate of the point where the corresponding terminal side of the angle intersects the unit circle and the **cosine ratio **\(\cost\) to bethe \(x\)-coordinate of the point where the corresponding terminal side of the angle intersects the unit circle. The**tangent ratio**\(\tan t\) is\(\dfrac{\sin t}{\cos t}\).

We may enclose the angle inparentheses or not depending on how clear the expression is:\(\sin t\) is the same as \(\sin (t)\) and \(\cos t\) is the same as \(\cos (t)\). When in doubt, use the extra parentheses when entering calculations into a calculator or computer.

SINE,COSINE, and TANGENTRATIOS

If \(t\) is an angle measurementand a point \((x,y)\) is both on the unit circle and the terminal side of the angle in standard position, then

\[ \begin{align} \cos t & = x \\ \sin t & = y \\ \tan t &=\dfrac{\sin t}{\cos t}\end{align}\]

Example \(\PageIndex{1}\)

Find \(\cos (90^{\text{o}})\) and \(\sin (90^{\text{o}}).\)

**Solution**

Moving \(90^{\text{o}}\) counterclockwise around the unit circle from the positive \(x\)-axis brings us to the top of the circle, where the \((x,y)\) coordinates are \((0,1)\), as shown here:

Using our definitions of cosine and sine,

\(\begin{align*} x &= \cos t = \cos (90^{\text{o}}) = 0 \\ y &= \sin t = \sin (90^{\text{o}}) = 1 \end{align*} \)

The cosine of \(90^{\text{o}}\) is 0; the sine of \(90^{\text{o}}\) is 1.

Try It \(\PageIndex{2}\)

Find cosine and sine of the angle \(\pi\).

**Answer**-
\(\cos (\pi)=−1, \sin (\pi)=0\)

### The Pythagorean Identity

Now that we can define sine and cosine, we will learn how they relate to each other and the unit circle. Recall that the equation for the unit circle is \(x^2+y^2=1\). Because \(x= \cos t\) and \(y=\sin t\), we can substitute for \( x\) and \(y\) to get \(\cos ^2 t+ \sin ^2 t=1.\) This equation, \( \cos ^2 t+ \sin ^2 t=1,\) is known as the **Pythagorean Identity**. See below:

We can use the Pythagorean Identity to find the cosine of an angle if we know the sine, or vice versa. However, because the equation yields two solutions, we need additional knowledge of the angle to choose the solution with the correct sign. If we know the quadrant where the angle is, we can easily choose the correct solution.

PYTHAGOREAN IDENTITY

The **Pythagorean Identity** states that, for any angle \(t\),

\[ \cos^2 t+ \sin^2 t=1 \]

How To: Given the sine of some angle t and its quadrant location, find the cosine of t

- Substitute the known value of \(\sin (t)\) into the Pythagorean Identity.
- Solve for \( \cos (t)\).
- Choose the solution with the appropriate sign for the \(x\)-values in the quadrant where \(t\) is located.

Example \(\PageIndex{3}\)

If \(\sin (t)=\dfrac{3}{7}\) and \(t\) is in the second quadrant, find \( \cos (t)\).

**Solution**

If we drop a vertical line from the point on the unit circle corresponding to \(t\), we create a right triangle, from which we can see that the Pythagorean Identity is simply one case of the Pythagorean Theorem.

Substituting the known value for sine into the Pythagorean Identity,

\[\begin{align*} \cos ^2 (t)+ \sin ^2(t) &=1 \\ \cos ^2(t)+\dfrac{9}{49} &=1 \\ \cos ^2(t) & = \dfrac{40}{49} \\ \cos (t) &=± \sqrt{\dfrac{40}{49}}=±\dfrac{\sqrt{40}}{7}=±\dfrac{2\sqrt{10}}{7} \end{align*}\]

Because the angle is in the second quadrant, we know the \(x\)-value is a negative real number, so the cosine is also negative. So

\[ \cos (t)=−\dfrac{2\sqrt{10}}{7} \nonumber \]

Try It \(\PageIndex{4}\)

If \(\cos (t)=\dfrac{24}{25}\) and \(t\) is in the fourth quadrant, find \( \sin (t)\).

**Answer**-
\(\sin (t)=−\dfrac{7}{25}\)

## Finding Sines,Cosinesand Tangent Ratios of Special Angles

When possible, we would like to be precise about evaluation of the sine and cosine ratios. We give an example here.

##### Example \(\PageIndex{5}\)

a) Find the value of \(\sin 150^\text{o}\)

b) Find the value of \(\cos\dfrac{7\pi}{4}\).

###### Solution

a) We first draw a picture and identify a triangle (while there are two natural possibilities, wewilluse the triangle shown below (often called "the reference triangle)).

Then determine the reference angle indicated in red. To do this we note that the straight angle (black angle plus the red angle is \(180^\text{o}\). So the red angle measures \(30^\text{o}\). Then we use our knowledge of the 30-60-90 triangle (choosing the one with hypotenuse 1) and note the lengths on the diagram.

Finally, making note of the position of the triangle, we see that the \(x\)-coordinate must be negative and the \(y\)-coordinate must be positive, we find

So we conclude, in particular, that \(\sin 150^\text{o}=\dfrac12\).

Note that we could also use the standard triangle (using the circle of radius 2)and use signed lengths and form the sine ratio as before:

b) We first draw the angle together with the unit circle and label the intersection:

We determine the value of the red angle (noting that the black angle plus the red angle measures \(2\pi\). Weconstruct a right triangle that will help us determine the coordinates of the labeledintersection (here we will use, again, what is often called 'the reference triangle').

Now, we recall the 45-45-90 triangle from memory (or rederive it) and scale it to have thehypotenusehave length1.

Now, we note from the location of the point that the \(x\)-coordinate must be positiveand the \(y\)-coordinate must be negative. We determine the coordinates of the labeled intersection.

It follows that \(\cos\dfrac{7\pi}{4}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\).

As in part a) we could also use the standard triangle (using the circle of radius 2)and use signed lengths and form the sine ratio as before:

To find a reference angle, draw a sketch noting in which quadrant your terminal side is and use your familiarity with right angles and straight angles to figure out the reference angle. Try these exercises using the above example as a model.

##### Try It \(\PageIndex{6}\)

Find the value of

a) \(\sin \dfrac{-5\pi}{3}\)

b) \(\cos 225^\text{o}\)

c) \(\tan 225^\text{o}\) (Here use the extension that \(\tan t=\dfrac{\sin t}{\cos t}.\)

**Answer**-
a) \(\dfrac{\sqrt{3}}{2}\)

b) \(\dfrac{-1}{\sqrt{2}}\)

c) 1

### Using a Calculator to Find Sine and Cosine

To find the cosine and sine of angles other than the **special angles**, we turn to a computer or calculator. **Be aware**: Most calculators can be set into “degree” or “radian” mode, which tells the calculator the units for the input value. When we evaluate \( \cos (30^\text{o})\) on our calculator, it will evaluate it as the cosine of 30 degrees if the calculator is in degree mode, or the cosine of 30 radians if the calculator is in radian mode.

Example \(\PageIndex{7}\): Using a Graphing Calculator to Find Sine and Cosine

a) Evaluate \( \sin150^\text{o}\) using a calculator or computer.

b) Evaluate \(\cos \dfrac{7\pi}{4}\)

**Solution**

a) This depends on your calculator. But we know the answer is \(\dfrac12\). Make sure the mode on the calculator is set to 'degree'.

b)This depends on your calculator. But we know the answer is \(\dfrac{1}{\sqrt{2}}\) so if you square your answer you should get \(\dfrac12\) and your answer should be positive. Make sure the mode on the calculator is set to 'radian'.

Try It \(\PageIndex{8}\)

Evaluate \(\sin(-50)\).

**Answer**-
Approximately -7.66

##### Writing Exercises \(\PageIndex{9}\)

- Explain how \(\cos(300{\text{o}})\) can be evaluated using the unit circle.
- Give four angles \(\theta, \alpha,\beta\) and \(\gamma\) for which \(\cos(300^{\text{o}})=\cos(\theta)=\cos(\alpha)=\cos(\beta)=\cos(\gamma)\)
- Explain what a reference triangle is and how you can create it. Is there another triangle that would give you the same benefit?
- If you know \(\sin(x)=\dfrac{4}{5}\), how can you find \(\cos(x)\) and \(\tan(x)\) without first finding \(x\). How might you use similar triangles/figures to make use of a circle of radius \(5\) instead of the unit circle?

##### Exit Problem

Evaluate \(\cos(-210^{\text{o}})\) and \(\sin(-210^{\text{o}})\). Support your answer with an appropriate picture of the unit circle. Then evaluate \(\tan(210^{\text{o}})\).

## Key Concepts

- Finding the function values for the sine and cosine begins with drawing a unit circle, which is centered at the origin and has a radius of 1 unit.
- Using the unit circle, the sine of an angle \(t\) equals the \(y\)-value of the endpoint on the unit circle of an arc of length \(t\) whereas the cosine of an angle \(t\) equals the \(x\)-value of the endpoint.
- The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis.
- When the sine or cosine is known, we can use the Pythagorean Identity to find the other. The Pythagorean Identity is also useful for determining the sines and cosines of special angles.
- Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known.
- The domain of the sine and cosine functions is all real numbers.
- The range of both the sine and cosine functions is \([−1,1]\).
- The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle.
- The signs of the sine and cosine are determined from the \(
*x\)*- and \(y\)-values in the quadrant of the original angle. - An angle’s reference angle is the size angle, \(t\), formed by the terminal side of the angle \(t\) and the horizontal axis.
- Reference angles can be used to find the sine and cosine of the original angle.
- Reference angles can also be used to find the coordinates of a point on a circle.