# Trigonometry

## Trigonometric equations and identities

### Part 1: Pythagorean identities

Recall that, in the section on the unit circle, we established that given any angle $\theta$, $\left(\cos\left(\theta\right),\sin\left(\theta\right)\right)$ are the coordinates of a point on the unit circle. Using the distance formula to compute the distance between this point and the origin, $\left(0,0\right)$, we get a trigonometric identity: $$\cos^2\left(\theta\right) + \sin^2\left(\theta\right) = 1$$
Dividing both sides by $\cos^2\left(\theta\right)$, we get another identity: $$1+\tan^2\left(\theta\right) = \sec^2\left(\theta\right)$$
Dividing both sides of the original identity by $\sin^2\left(\theta\right)$, we get the identity: $$\cot^2\left(\theta\right)+1=\csc^2\left(\theta\right)$$

### Example. Prove the following identity: $\frac{\sin\left(x\right)}{1+\cos\left(x\right)}=\frac{1-\cos\left(x\right)}{\sin\left(x\right)}$

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

### Part 2: Even-odd identities

We also have the following identities: $$\sin\left(-\theta\right)=-\sin\left(\theta\right) \hspace{10 mm} \cos\left(-\theta\right)=\cos\left(\theta\right) \hspace{10 mm} \tan\left(-\theta\right)=-\tan\left(\theta\right)$$ They tell us that $\sin$ and $\tan$ are oddAn odd function $f$ is any function which satisfies $f(-x) = -f(x)$ for all $x$ in its domain. functions while $\cos$ is an evenAn even function $f$ is any function which satisfies $f(-x) = f(x)$ for all $x$ in its domain. function.

### Example. Compute $\sin\left(-\frac{\pi}{2}\right)$

$$\sin\left(-\frac{\pi}{2}\right)=-\sin\left(\frac{\pi}{2}\right)=-1$$

We can reduce $\sin$ and $\cos$ of the sum of two angles to an expression involving $\sin$ and $\cos$ of each of the angles: \begin{align*} \sin\left(A+B\right) &= \sin\left(A\right)\cos\left(B\right) + \cos\left(A\right)\sin\left(B\right) \\ \cos\left(A+B\right) &= \cos\left(A\right)\cos\left(B\right) - \sin\left(A\right)\sin\left(B\right) \end{align*}

### Example. Evaluate $\sin\left(75^{\circ}\right)$.

\begin{align*} \sin\left(75^{\circ}\right) &= \sin\left(45^{\circ}+30^{\circ}\right) \\ &= \sin\left(45^{\circ}\right)\cos\left(30^{\circ}\right) + \cos\left(45^{\circ}\right)\sin\left(30^{\circ}\right) \\ &= \frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{\sqrt{2}}\left(\frac{1}{2}\right) \\ &= \frac{\sqrt{3}+1}{2\sqrt{2}} \end{align*}

### Part 4: Trigonometric equations

The techniques for solving trigonometric equations involve the same strategies as solving polynomial equations (see the section on Polynomials and Factoring) as well as using trigonometric identities.

### Example. Find the solutions of the equation

$$\left(\sin\left(\theta\right) - 4\right) \left(\sin^2\left(\theta\right)+5\sin\left(\theta\right)+4\right) = 0 \qquad \textrm{ for } \theta \in \left[0,2\pi \right].$$
We need to find when the first or the second factor is zero. Since $\sin\left(\theta\right) - 4=0$ means $\sin\left(\theta\right)=4$, which has no solution (the left side is at most one), we only need to work with the second factor: $$\sin^2\left(\theta\right)+5\sin\left(\theta\right)+4 = 0$$ Notice that this is a quadratic equation, and after factoring we get: $$(\sin\left(\theta\right)+1)(\sin\left(\theta\right)+4) = 0$$ Again, we need to check when either factor is zero. For the second factor, we get $\sin\left(\theta\right)+4 = 0$, i.e. $\sin\left(\theta\right)=-4$, which has no solution (since the left side is at least $-1$). Solving for the first factor, we get $\sin\left(\theta\right)+1=0$, i.e. $\sin\left(\theta\right)=-1$. The only $\theta \in \left[0,2\pi \right]$ for which this is true is $\theta= \frac{3\pi}{2}$.