Chapter 7, Problem 27CT

To determine

The solution of equation ${\cos }^2\theta +2\sin \theta \cos \theta -{\sin }^2\theta =0$ on $0\le \theta <2\pi$.

Answer to Problem 27CT

Solution:

The solution of equation ${\cos }^2\theta +2\sin \theta \cos \theta -{\sin }^2\theta =0$ on $0\le \theta <2\pi$ is $\theta =\frac{3\pi }{8},\frac{7\pi }{8},\frac{11\pi }{8}$ and $\frac{15\pi }{8}$

Explanation of Solution

Given information:

The equation ${\cos }^2\theta +2\sin \theta \cos \theta -{\sin }^2\theta =0$

Explanation:

Consider the equation ${\cos }^2\theta +2\sin \theta \cos \theta -{\sin }^2\theta =0$

$\Rightarrow \left({\cos }^2\theta -{\sin }^2\theta \right)+2\sin \theta \cos \theta =0$

By using ${\cos }^2\theta -{\sin }^2\theta =\cos \left(2\theta \right)$ and $2\sin \theta \cos \theta =\sin \left(2\theta \right)$,

$\Rightarrow \cos \left(2\theta \right)+\sin \left(2\theta \right)=0$

Subtracting $\sin \left(2\theta \right)$ from both sides,

$\Rightarrow \cos \left(2\theta \right)=-\sin \left(2\theta \right)$

Dividing both sides by $\cos \left(2\theta \right)$

$\Rightarrow \frac{\cos \left(2\theta \right)}{\cos \left(2\theta \right)}=-\frac{\sin \left(2\theta \right)}{\cos \left(2\theta \right)}$

By using $\frac{\sin u}{\cos u}=\tan u$,

$\Rightarrow \tan \left(2\theta \right)=-1$

The tan is negative in II and III quadrant.

The argument is $2\theta$

Write the general formula that gives all the solutions of the equation $\tan \left(2\theta \right)=-1$, then list those that are in the interval $[0,2\pi )$

Since $\tan \left(\frac{3\pi }{4}+k\pi \right)=-1$ for any integer $k$,

$\Rightarrow 2\theta =\frac{3\pi }{4}+k\pi$

$\Rightarrow \theta =\frac{3\pi }{8}+\frac{k\pi }{2}$

$\Rightarrow \theta =\frac{\pi }{8}\left(3+4k\right)$

If $k=0$, then $\theta =\frac{\pi }{8}\left(3+4\left(0\right)\right)=\frac{3\pi }{8}$

If $k=1$, then $\theta =\frac{\pi }{8}\left(3+4\left(1\right)\right)=\frac{7\pi }{8}$

If $k=2$, then $\theta =\frac{\pi }{8}\left(3+4\left(2\right)\right)=\frac{11\pi }{8}$

If $k=3$, then $\theta =\frac{\pi }{8}\left(3+4\left(3\right)\right)=\frac{15\pi }{8}$

Therefore, the solution of equation ${\cos }^2\theta +2\sin \theta \cos \theta -{\sin }^2\theta =0$ on $0\le \theta <2\pi$ are $\theta =\frac{3\pi }{8},\frac{7\pi }{8},\frac{11\pi }{8}$ and $\frac{15\pi }{8}$.