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

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

Part 3: Addition formulas

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 $\theta= \frac{3\pi}{2}$.