## Trigonometry (11th Edition) Clone

The solution set to this problem is $$\{45^\circ, 135^\circ, 225^\circ, 315^\circ\}$$
$$\cos^2\theta-\sin^2\theta=0$$ over interval $[0^\circ,360^\circ)$ 1) Solve the equation: $$\cos^2\theta-\sin^2\theta=0$$ - Recall the identity: $\sin^2\theta=1-\cos^2\theta$ $$\cos^2\theta-(1-\cos^2\theta)=0$$ $$\cos^2\theta-1+\cos^2\theta=0$$ $$2\cos^2\theta-1=0$$ $$\cos^2\theta=\frac{1}{2}$$ $$\cos\theta=\pm\frac{1}{\sqrt2}=\pm\frac{\sqrt2}{2}$$ 2) Apply the inverse function: - For $\cos\theta=\frac{\sqrt2}{2}$: Over the interval $[0^\circ,360^\circ)$, $\cos\theta\gt0$ means the value of $\theta$ lies either in quadrant I or IV. In fact, the angles of $45^\circ$ in quadrant I and $315^\circ$ in quadrant IV are where cosine equals $\frac{\sqrt2}{2}$. - For $\cos\theta=-\frac{\sqrt2}{2}$: Over the interval $[0^\circ,360^\circ)$, $\cos\theta\lt0$ means the value of $\theta$ lies either in quadrant II or III. In fact, the angles of $135^\circ$ in quadrant II and $225^\circ$ in quadrant III are where cosine equals $-\frac{\sqrt2}{2}$. Therefore, $$\theta\in\{45^\circ, 135^\circ, 225^\circ, 315^\circ\}$$ In other words, the solution set to this problem is $$\{45^\circ, 135^\circ, 225^\circ, 315^\circ\}$$