solving trig equations | Racquel Sanderson

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August 10, 2025 · 4:54 am

Trigonometric Identities – Product to Sum

Let’s think about the sine and cosine addition and subtraction trig identities.

(1) $\begin{equation*}sin(A+B)=sinAcosB+cosAsinB\end{equation*}$

(2) $\begin{equation*}sin(A-B)=sinAcosB-cosAsinB\end{equation*}$

If we add equation $1$ and $2$ , we get

$\begin{equation*}sin(A+B)+sin(A-B)=2sinAcosB\end{equation}$

Hence, $sinAcosB=\frac{1}{2}(sin(A+B)+sin(A-B))$

If we subtract equation $2$ from equation $1$ , we get

$\begin{equation*}sin(A+B)-sin(A-B)=2cosAsinB\end{equation}$

Hence, $cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B)$

What about the cosine addition and subtraction idenities?

(3) $\begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation*}$

(4) $\begin{equation*}cos(A-B)=cosAcosB+sinAsinB\end{equation*}$

If we add equation $3$ and $4$ , we get

$\begin{equation*}cos(A+B)+cos(A-B)=2cosAcosB\end{equation}$

Hence, $cosAcosB=\frac{1}{2}(cos(A+B)+cos(A-B))$

If we subtract $3$ from $4$ , we get

$\begin{equation*}cos(A-B)-cos(A+B)=2sinAsinB\end{equation}$

Hence, $sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

These are the product to sum identities.

$\begin{equation*}sinAcosB=\frac{1}{2}(sin(A+B)+sin(A-B))\end{equation}$

$\begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}$

$\begin{equation*}cosAcosB=\frac{1}{2}(cos(A+B)+cos(A-B))\end{equation}$

$\begin{equation*}sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))\end{equation}$

Examples

(1) Solve $sin(5x)-sin(x)=0$ for $0\le x \le 2\pi$

Remember,

$\begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}$

$\begin{equation*}A+B=5\end{equation}$

$\begin{equation*}A-B=1\end{equation}$

Therefore, $A=3$ and $B=2$

$\begin{equation*}sin(5x)-sin(x)=2cos(3x)sin(2x)=0\end{equation}$

$\begin{equation*}cos(3x)=0\end{equation}$

$\begin{equation*}3x=\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{\9\pi}{2}, \frac{11\pi}{2}\end{equation}$

$\begin{equation*}x=\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}\end{equation}$

$\begin{equation*}sin(2x)=0\end{equation}$

$\begin{equation*}2x=0, \pi, 2\pi, 3\pi, 4\pi\end{equation}$

$\begin{equation*}x=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\end{equation}$

Hence $x=0, \frac{\pi}{6}. \frac{\pi}{2}, \frac{5\pi}{6}, \pi, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}, 2\pi$

(2)Solve $sin(7\theta)-sin(\theta)=sin(3\theta)$ for $0 \le \theta \le\2\pi$

$\begin{equation*}cosAsinB=\frac{1}{2}(sin(A+B)-sin(A-B))\end{equation}$

Therefore, $A+B=7$ and $A-B=1$

$A=4, B=3$

$\begin{equation*}2cos(4\theta)sin(3\theta)=sin(3\theta)\end{equation}$

$\begin{equation*}2cos(4\theta)sin(3\theta)-sin(3\theta)=0\end{equation}$

$\begin{equation*}sin(3\theta)(2cos(4\theta)-1)=0\end{equation}$

$sin(3\theta)=0$ and $cos(4\theta)=\frac{1}{2}$

$3\theta=0, \pi, 2\pi, 3\pi, 4\pi, 5\pi, 6\pi$

$\theta=0, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}. 2\pi$

$cos(4\theta)=\frac{1}{2}$

$4\theta=\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}, \frac{13\pi}{3}, \frac{17\pi}{3}, \frac{19\pi}{3}, \frac{23\pi}{3}$

$\theta=\frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{19\pi}{12}, \frac{23\pi}{12}$

Hence $\theta=0, \frac{\pi}{3}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}, \frac{5\pi}{3}, \frac{23\pi}{12}, 2\pi$

General Solutions to Trigonometric Equations

Solve $sinx=\frac{1}{2}$ for $0\le x \le 2\pi$

Sine is positive in the first and second quadrants.

$\begin{equation*}sinx=\frac{1}{2}\end{equation}$

$\begin{equation*}x=\frac{\pi}{6} \text{ and } x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\end{equation}$

But what if we aren’t given a domain for the $x$ values?

Then we need to give general solutions.

For example,

Solve $sinx=\frac{1}{2}$

As you can see from the sketch above, there are infinite solutions.

The sine function has a period of $360^\circ$ , and so if $\frac{\pi}{6}$ is a solution then $\2pi+\frac{\pi}{6}$ is also a solution. This means $\frac{\pi}{6}+2\pi n, n\in\mathbb{Z}$ is a general solution. And we can do the same for the second solution $\frac{5\pi}{6}+2\pi n$ .

In general

$\begin{equation*}sinx=y\end{equation}$

$\begin{equation*}x=arcsin(y)+2\pi n \text { and } x=\pi-arcsin(y)+2\pi n \end{equation}$

$\begin{equation*}x=arcsin(y)+2\pi n \text { and } x=\pi(2n+1)-arcsin(y), n \in \mathbb{Z}\end{equation}$

We can turn this into one equation
$\begin{equation*}x=(-1)^n arcsin(y)+n\pi, n \in \mathbb{Z}\end{equation}$

What about cosine?

Solve $cosx=\frac{1}{2}$

Cosine is positive in the first and fourth quadrants (it also has a period of $2\pi$ . The first two (positive) solutions are $\frac{\pi}{3}$ and $2\pi-\frac{\pi}{3}$ .