## Trigonometry (11th Edition) Clone

$$\sin\Big(\cos^{-1}(-\frac{1}{2})+\tan^{-1}(-\sqrt3)\Big)=-\frac{\sqrt3}{2}$$
$$\sin\Big(\cos^{-1}(-\frac{1}{2})+\tan^{-1}(-\sqrt3)\Big)$$ 1) $\cos^{-1}(-\frac{1}{2})$ $-\frac{1}{2}\lt0$ means that you can find the answer in either quadrant II or III, the area of negativity for a cosine function. In quadrant II, there is this value of $\frac{2\pi}{3}$ that $\cos\frac{2\pi}{3}=-\frac{1}{2}$. So, $$\cos^{-1}(-\frac{1}{2})=\frac{2\pi}{3}$$ 2) $\tan^{-1}(-\sqrt3)$ $-\sqrt3\lt0$ means that you can find the answer in either quadrant II or IV, the area of negativity for a tangent function. In quadrant II, $\frac{2\pi}{3}$ is also the angle that $\tan\frac{2\pi}{3}=-\sqrt3$. Therefore, $$\tan^{-1}(-\sqrt3)=\frac{2\pi}{3}$$ 3) Refer back to the original formula $$\sin\Big(\cos^{-1}(-\frac{1}{2})+\tan^{-1}(-\sqrt3)\Big)$$ $$=\sin\Big(\frac{2\pi}{3}+\frac{2\pi}{3}\Big)$$ $$=\sin\frac{4\pi}{3}$$ $\frac{4\pi}{3}$ lies in quadrant III, and corresponds to angle $\frac{\pi}{3}$ in quadrant I. However, quadrant III is where $\sin\theta\lt0$. Therefore, $$=-\sin\frac{\pi}{3}$$ $$=-\frac{\sqrt3}{2}$$