You are given a list of Trig Identities. One of those identities is \begin{align*}\cos (- \theta)=\cos \theta\end{align*}. Prove this identity without graphing.

### Trigonometric Identities

**Trigonometric identities** are true for any value of \begin{align*}x\end{align*} (as long as the value is in the domain). You have learned about secant, cosecant, and cotangent, which are all reciprocal functions of sine, cosine and tangent. These functions can be rewritten as the Reciprocal Identities because they are always true.

**Reciprocal Identities:** \begin{align*}\csc \theta=\frac{1}{\sin \theta} \quad \sec \theta=\frac{1}{\cos \theta} \quad \cot \theta=\frac{1}{\tan \theta}\end{align*}

Other identities involve the tangent, variations on the Pythagorean Theorem, phase shifts, and negative angles. We will discover them in this concept.

We know that \begin{align*}\tan \theta=\frac{opposite}{adjacent}\end{align*}. Let's show that \begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*}. This is the **Tangent Identity**.

Whenever we are trying to verify, or prove, an identity, we start with the statement we are trying to prove and work towards the desired answer. In this case, we will start with \begin{align*}\tan \theta=\frac{\sin \theta}{\cos \theta}\end{align*} and show that it is equivalent to \begin{align*}\tan \theta=\frac{opposite}{adjacent}\end{align*}. First, rewrite sine and cosine in terms of the ratios of the sides.

\begin{align*}\tan \theta&=\frac{\sin \theta}{\cos \theta} \\ &=\frac{\frac{opposite}{hypotenuse}}{\frac{adjacent}{hypotenuse}}\end{align*}

Then, rewrite the complex fraction as a division problem and simplify.

\begin{align*}&=\frac{opposite}{hypotenuse} \div \frac{adjacent}{hypotenuse} \\ &=\frac{opposite}{\cancel{hypotenuse}} \cdot \frac{\cancel{hypotenuse}}{adjacent} \\ &=\frac{opposite}{adjacent}\end{align*}

We now have what we wanted to prove and we are done. **Once you verify an identity, you may use it to verify other identities.**

Now, let's show that \begin{align*}\sin^2 \theta+ \cos^2 \theta=1\end{align*} is a true identity.

Change the sine and cosine in the equation into the ratios. In this problem, we will use \begin{align*}y\end{align*} as the opposite side, \begin{align*}x\end{align*} is the adjacent side, and \begin{align*}r\end{align*} is the hypotenuse (or radius), as in the unit circle.

\begin{align*}\sin^2 \theta+ \cos^2 \theta&=1 \\ \left(\frac{y}{r}\right)^2+\left(\frac{x}{r}\right)^2&=1 \\ \frac{y^2}{r^2}+\frac{x^2}{r^2}&=1 \\ \frac{y^2+x^2}{r^2}&=1\end{align*}

Now, \begin{align*}x^2+y^2=r^2\end{align*} from the Pythagorean Theorem. Substitute this in for the numerator of the fraction.

\begin{align*}\frac{r^2}{r^2}=1\end{align*}

This is one of the **Pythagorean Identities** and very useful.

Finally, let's verify that \begin{align*}\sin \left(\frac{\pi}{2}- \theta \right)=\cos \theta\end{align*} by using the graphs of the functions.

The function \begin{align*}y=\sin \left(\frac{\pi}{2} - x \right)\end{align*} is a phase shift of \begin{align*}\frac{\pi}{2}\end{align*} of the sine curve.

The red function above is \begin{align*}y=\sin x\end{align*} and the blue is \begin{align*}y=\cos x\end{align*}. If we were to shift the sine curve \begin{align*}\frac{\pi}{2}\end{align*}, it would overlap perfectly with the cosine curve, thus proving this **Cofunction Identity**.

### Examples

#### Example 1

Earlier, you were asked to prove the identity of \begin{align*}\cos (- \theta)=\cos \theta\end{align*} without graphing.

First, recall that \begin{align*}\cos\theta=x\end{align*}, where \begin{align*}(x, y)\end{align*} is the endpoint of the terminal side of \begin{align*}\theta\end{align*} on the unit circle.

Now, if we have \begin{align*}\cos (- \theta)\end{align*}, what is its endpoint? Well, the negative sign tells us that the angle is rotated in a clockwise direction, rather than the usual counter-clockwise. If we make this rotation, we see that \begin{align*}\cos (- \theta)=x\end{align*} as well, as illustrated in the following diagram.

We know that \begin{align*}x = x\end{align*}, so we can set the two expressions equal to one another.

\begin{align*}\cos\theta = \cos (- \theta) \end{align*}

We can now flip this identity around to get:

\begin{align*}\cos (- \theta)=\cos \theta\end{align*}

#### Example 2

Prove the Pythagorean Identity: \begin{align*}1+ \tan^2 \theta=\sec^2 \theta\end{align*}.

First, let’s use the Tangent Identity and the Reciprocal Identity to change tangent and secant in terms of sine and cosine.

\begin{align*}1+ \tan^2 \theta&=\sec^2 \theta \\ 1+ \frac{\sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta}\end{align*}

Now, change the 1 into a fraction with a base of \begin{align*}\cos^2 \theta\end{align*} and simplify.

\begin{align*}1+ \frac{\sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta} \\ \frac{\cos^2 \theta}{\cos^2 \theta}+ \frac{\sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta} \\ \frac{\cos^2 \theta+ \sin^2 \theta}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta} \\ \frac{1}{\cos^2 \theta}&=\frac{1}{\cos^2 \theta}\end{align*}

In the second to last step, we arrived at the original Pythagorean Identity \begin{align*}\sin^2 \theta+\cos^2 \theta\end{align*} in the numerator of the left-hand side. Therefore, we can substitute in 1 for this and the two sides of the equation are the same.

#### Example 3

Without graphing, show that \begin{align*}\sin (- \theta)=- \sin \theta\end{align*}.

First, recall that \begin{align*}\sin \theta=y\end{align*}, where \begin{align*}(x, y)\end{align*} is the endpoint of the terminal side of \begin{align*}\theta\end{align*} on the unit circle.

Now, if we have \begin{align*}\sin (- \theta)\end{align*}, what is it’s endpoint? Well, the negative sign tells us that the angle is rotated in a clockwise direction, rather than the usual counter-clockwise. By looking at the picture, we see that \begin{align*}\sin (- \theta)=-y\end{align*}. Therefore, if \begin{align*}\sin \theta=y\end{align*}, then \begin{align*}- \sin \theta=-y\end{align*} and combining the equations, we have \begin{align*}\sin (- \theta)=- \sin \theta\end{align*}.

### Review

- Show that \begin{align*}\cot \theta=\frac{\cos \theta}{\sin \theta}\end{align*}.
- Show that \begin{align*}\tan \theta=\frac{\sec \theta}{\csc \theta}\end{align*}.
- Show that \begin{align*}1+ \cot^2 \theta=\csc^2 \theta\end{align*}.
- Explain why \begin{align*}\cos \left(\frac{\pi}{2} - \theta \right)=\sin \theta\end{align*} by using the graphs of the two functions.
- Show that \begin{align*}\sec (- \theta)=\sec \theta\end{align*}.
- Explain why \begin{align*}\tan (- \theta)=- \tan \theta\end{align*} is true, using the Tangent Identity and the other Negative Angle Identities.

Verify the following identities.

- \begin{align*}\cot \theta \sec \theta=\csc \theta\end{align*}
- \begin{align*}\frac{\cos \theta}{\cot \theta}=\frac{\tan \theta}{\sec \theta}\end{align*}
- \begin{align*}\sin \theta \csc \theta=1\end{align*}
- \begin{align*}\cot(- \theta)=- \cot \theta\end{align*}
- \begin{align*}\tan x \csc x \cos x=1\end{align*}
- \begin{align*}\frac{\sin^2 \left(-x\right)}{\tan^2 x}= \cos ^2 x\end{align*}

Show that \begin{align*}\sin^2 \theta+ \cos^2 \theta=1\end{align*} is true for the following values of \begin{align*}\theta\end{align*}.

- \begin{align*}\frac{\pi}{4}\end{align*}
- \begin{align*}\frac{2 \pi}{3}\end{align*}
- \begin{align*}- \frac{7 \pi}{6}\end{align*}
- Recall that a function is odd if \begin{align*}f(-x)=-f(x)\end{align*} and even if \begin{align*}f(-x)=f(x)\end{align*}. Which of the six trigonometric functions are odd? Which are even?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.6.