<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation
Our Terms of Use (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use.

Solving Trigonometric Equations using Sum and Difference Formulas

Solve sine, cosine, and tangent of angles that are added or subtracted.

Atoms Practice
Estimated10 minsto complete
Practice Solving Trigonometric Equations using Sum and Difference Formulas
This indicates how strong in your memory this concept is
Estimated10 minsto complete
Practice Now
Turn In
Solving Trig Equations using Sum and Difference Formulas

As Agent Trigonometry, you are given a piece of the puzzle: \begin{align*}\sin(\frac{\pi}{2} - x) = -1\end{align*}. What is the value of \begin{align*}x\end{align*}?

Solving Trigonometric Functions

We can use the sum and difference formulas to solve trigonometric equations. For this concept, we will only find solutions in the interval \begin{align*}0\le x <2\pi\end{align*}.

Let's solve the following functions using the sum and difference formulas.

  1. \begin{align*}\cos (x-\pi)=\frac{\sqrt{2}}{2}\end{align*}

Use the formula to simplify the left-hand side and then solve for \begin{align*}x\end{align*}.

\begin{align*}\cos (x-\pi) &=\frac{\sqrt{2}}{2} \\ \cos x \cos \pi +\sin x \sin \pi &=\frac{\sqrt{2}}{2} \\ -\cos x &=\frac{\sqrt{2}}{2}\\ \cos x &=-\frac{\sqrt{2}}{2}\end{align*}

The cosine negative in the \begin{align*}2^{nd}\end{align*} and \begin{align*}3^{rd}\end{align*} quadrants. \begin{align*}x=\frac{3\pi}{4}\end{align*} and \begin{align*}\frac{5\pi}{4}\end{align*}.

  1. \begin{align*}\sin \left(x+\frac{\pi}{4}\right)+1=\sin \left(\frac{\pi}{4}-x\right)\end{align*}

\begin{align*}\sin \left(x+\frac{\pi}{4}\right)+1 &=\sin \left(\frac{\pi}{4}-x\right) \\ \sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4}+1 &=\sin \frac{\pi}{4}\cos x -\cos \frac{\pi}{4}\sin x \\ \sin x \cdot \frac{\sqrt{2}}{2}+\cos x \cdot \frac{\sqrt{2}}{2}+1 &=\frac{\sqrt{2}}{2}.\cos x -\frac{\sqrt{2}}{2} \cdot \sin x \\ \sqrt{2} \sin x &=-1 \\ \sin x &=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\end{align*}

In the interval, \begin{align*}x=\frac{5\pi}{4}\end{align*} and \begin{align*}\frac{7\pi}{4}\end{align*}.

  1. \begin{align*}2\sin \left(x+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}\end{align*}

\begin{align*}2\sin\left(x+\frac{\pi}{3}\right) &=\tan \frac{\pi}{3} \\ 2\left(\sin x \cos \frac{\pi}{3}+\cos x \sin \frac{\pi}{3}\right) &=\sqrt{3} \\ 2\sin x \cdot \frac{1}{2}+2\cos x \cdot \frac{\sqrt{3}}{2} &=\sqrt{3} \\ \sin x +\sqrt{3}\cos x &=\sqrt {3} \\ \sin x &=\sqrt{3}(1-\cos x) \\ \sin ^2x &=3(1-2\cos x+\cos ^2x) \qquad \text{square both sides} \\ 1-\cos ^2x &=3-6\cos x + 3\cos ^2 x \qquad \ \ \text{substitute} \ \sin ^2{x} = 1-\cos ^2x \\ 0 &=4\cos^2x-6\cos x +2 \\ 0 &=2\cos^2x-3\cos x +1\end{align*}

At this point, we can factor the equation to be \begin{align*}(2\cos x -1)(\cos x -1)=0\end{align*}. \begin{align*}\cos x =\frac{1}{2}\end{align*}, and 1, so \begin{align*}x=0, \frac{\pi}{3}, \frac{5 \pi}{3}\end{align*}. Be careful with these answers. When we check these solutions it turns out that \begin{align*}\frac{5\pi}{3}\end{align*} does not work.

\begin{align*}2\sin\left(\frac{5\pi}{3}+\frac{\pi}{3}\right) &=\tan \frac{\pi}{3} \\ 2\sin 2\pi &=\sqrt{3} \\ 0 &\ne \sqrt{3}\end{align*}

Therefore, \begin{align*}\frac{5\pi}{3}\end{align*} is an extraneous solution.


Example 1

Earlier, you were asked to find the value of x from the equation \begin{align*}\sin(\frac{\pi}{2} - x) = -1\end{align*}

First, simplify the expression \begin{align*}\sin(\frac{\pi}{2} - x)\end{align*} as:

\begin{align*}\sin(\frac{\pi}{2} - x)=\sin \frac{\pi}{2} \cos x - \cos \frac{\pi}{2} \sin x \\ &=1\cdot \cos x - 0\cdot \sin x \\ &=cos x\end{align*}

So what you're now looking for is the value of \begin{align*}x\end{align*} where \begin{align*}\cos x = -1\end{align*}.

The cosine of \begin{align*}180^\circ\end{align*} is equal to \begin{align*}-1\end{align*}.

Solve the following equations in the interval \begin{align*}0\le x<2\pi\end{align*}.

Example 2

\begin{align*}\cos(2\pi - x)=\frac{1}{2}\end{align*}
\begin{align*}\cos (2\pi-x) &=\frac{1}{2} \\ \cos 2\pi \cos x +\sin 2 \pi \sin x &=\frac{1}{2} \\ \cos x &=\frac{1}{2} \\ x &=\frac{\pi}{3} \ and \ \frac{5\pi}{3}\end{align*}

Example 3

\begin{align*}\sin \left(\frac{\pi}{6}-x\right)+1 = \sin \left(x+\frac{\pi}{6}\right)\end{align*}

\begin{align*}\sin \left(\frac{\pi}{6}-x\right)+1 &=\sin \left(x+\frac{\pi}{6}\right)\\ \sin \frac{\pi}{6}\cos x-\cos \frac{\pi}{6}\sin x +1 &=\sin x \cos \frac{\pi}{6}+\cos x \sin \frac{\pi}{6} \\ \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x+1 &=\frac{\sqrt{3}}{2}\sin x +\frac{1}{2} \cos x \\ 1 &=\sqrt{3}\sin x \\ \frac{1}{\sqrt{3}} &=\sin x \\ x &=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)=0.6155 \ and \ 2.5261 \ \text{rad}\end{align*}

Example 4

\begin{align*}\cos \left(\frac{\pi}{2}+x\right)=\tan \frac{\pi}{4}\end{align*}
\begin{align*}\cos \left(\frac{\pi}{2}+x\right) &=\tan \frac{\pi}{4} \\ \cos \frac{\pi}{2} \cos x - \sin \frac{\pi}{2} \sin x &=1 \\ -\sin x &=1 \\ \sin x &=-1 \\ x &=\frac{3\pi}{2}\end{align*}


Solve the following trig equations in the interval \begin{align*}0\le x < 2\pi\end{align*}.

  1. \begin{align*}\sin (x-\pi)=-\frac{\sqrt{2}}{2}\end{align*}
  2. \begin{align*}\cos(2\pi +x)=-1\end{align*}
  3. \begin{align*}\tan \left(x+\frac{\pi}{4}\right)=1\end{align*}
  4. \begin{align*}\sin \left(\frac{\pi}{2}-x\right)=\frac{1}{2}\end{align*}
  5. \begin{align*}\sin \left(x+\frac{3\pi}{4}\right)+\sin \left(x-\frac{3\pi}{4}\right)=1\end{align*}
  6. \begin{align*}\sin \left(x+\frac{\pi}{6}\right)=-\sin \left(x-\frac{\pi}{6}\right)\end{align*}
  7. \begin{align*}\cos \left(x+\frac{\pi}{6}\right)=\cos \left(x-\frac{\pi}{6}\right)+1\end{align*}
  8. \begin{align*}\cos \left(x+\frac{\pi}{3}\right)+\cos \left(x-\frac{\pi}{3}\right)=1\end{align*}
  9. \begin{align*}\tan(x+\pi)+2\sin (x+\pi)=0\end{align*}
  10. \begin{align*}\tan (x+\pi)+\cos \left(x+\frac{\pi}{2}\right)=0\end{align*}
  11. \begin{align*}\tan \left(x+\frac{\pi}{4}\right)=\tan \left(x-\frac{\pi}{4}\right)\end{align*}
  12. \begin{align*}\sin \left(x-\frac{5\pi}{3}\right)-\sin \left(x-\frac{2\pi}{3}\right)=0\end{align*}
  13. \begin{align*}4\sin (x+\pi)-2=2\cos\left(x+\frac{\pi}{2}\right)\end{align*}
  14. \begin{align*}1+2\cos(x-\pi)+\cos x =0\end{align*}
  15. Real Life Application The height, \begin{align*}h\end{align*} (in feet), of two people in different seats on a Ferris wheel can be modeled by \begin{align*}h_1=50\cos 3t+46\end{align*} and \begin{align*}h_2=50\cos 3\left(t-\frac{3\pi}{4}\right)+46\end{align*} where \begin{align*}t\end{align*} is the time (in minutes). When are the two people at the same height?

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 14.14. 

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Solving Trigonometric Equations using Sum and Difference Formulas.
Please wait...
Please wait...