Question 21463

Nov 30, 2016

For $0 \le \theta < 2 \pi$, $\theta = \frac{\pi}{5} , \frac{5 \pi}{6} , \frac{3 \pi}{2}$

Explanation:

${\cos}^{2} \theta - {\sin}^{2} \theta = \sin \theta$

Use the trig identity ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$
$\implies {\cos}^{2} \theta = 1 - {\sin}^{2} \theta$

Substitute for ${\cos}^{2} \theta$

$1 - {\sin}^{2} \theta - {\sin}^{2} \theta = \sin \theta$

$- 2 {\sin}^{2} \theta + 1 = \sin \theta$

$0 = 2 {\sin}^{2} \theta + \sin \theta - 1$

Factor

$\left(2 \sin \theta - 1\right) \left(\sin \theta + 1\right) = 0$

Set each factor equal to zero and solve.

$2 \sin \theta - 1 = 0 \textcolor{w h i t e}{a a a} \sin \theta + 1 = 0$

$\sin \theta = \frac{1}{2} \textcolor{w h i t e}{a a a} \sin \theta = - 1$

For $0 \le \theta < 2 \pi$, use unit circle to find

$\theta = \frac{\pi}{6} , \frac{5 \pi}{6} \textcolor{w h i t e}{a a a} \theta = \frac{3 \pi}{2}$

Nov 30, 2016

${\cos}^{2} \theta - {\sin}^{2} \theta = \sin \theta$

$\implies \cos 2 \theta = \cos \left(\frac{\pi}{2} - \theta\right)$

2theta=2npi±(pi/2-theta)," where "n in ZZ#

When $2 \theta = 2 n \pi + \left(\frac{\pi}{2} - \theta\right)$

$\implies 3 \theta = \left(4 n + 1\right) \frac{\pi}{2}$

$\implies \theta = \left(4 n + 1\right) \frac{\pi}{6}$

Again when

$2 \theta = 2 n \pi - \left(\frac{\pi}{2} - \theta\right)$

$\implies 2 \theta = 2 n \pi - \frac{\pi}{2} + \theta$

$\implies \theta = \left(4 n - 1\right) \frac{\pi}{2}$