## Trigonometry (11th Edition) Clone

$$\sin2x=\frac{15}{17}$$ $$\cos2x=-\frac{8}{17}$$
$$\tan x=\frac{5}{3} \hspace{2cm}\sin x\lt0$$ $$\sin 2x=?\hspace{2cm}\cos2x=?$$ 1) First, we need to calculate $\sin x$ and $\cos x$ as they are essential to the calculations of $\sin 2x$ and $\cos 2x$. We know that $\tan x=\frac{\sin x}{\cos x}$. Thus, since $\tan x=\frac{5}{3}\gt0$ and $\sin x\lt0$, we deduce that $\cos x\lt0$. According to Pythagorean Identities: $$\tan^2x+1=\sec^2x$$ According to Reciprocal Identities: $$\sec x=\frac{1}{\cos x}$$ Therefore, $$\tan^2x+1=\frac{1}{\cos^2x}$$ $$\cos^2x=\frac{1}{\tan^2x+1}=\frac{1}{\Big(\frac{5}{3}\Big)^2+1}=\frac{1}{\frac{25}{9}+1}=\frac{1}{\frac{34}{9}}=\frac{9}{34}$$ $$\cos x=-\frac{3}{\sqrt{34}}=-\frac{3\sqrt{34}}{34}\hspace{1cm}(\cos x\lt0)$$ Now we find $\sin x$, using Quotient Identities: $$\tan x=\frac{\sin x}{\cos x}$$ $$\sin x=\tan x\cos x=\frac{5}{3}\times\Big(-\frac{3\sqrt{34}}{34}\Big)=-\frac{5\sqrt{34}}{34}$$ 2) Now we can calculate $\sin2x$ and $\cos2x$ using Double-Angle Identities $$\sin2x=2\sin x\cos x=2\times\Big(-\frac{5\sqrt{34}}{34}\Big)\times\Big(-\frac{3\sqrt{34}}{34}\Big)=\frac{2\times15\times34}{34^2}=\frac{15}{17}$$ $$\cos 2x=\cos^2x-\sin^2x=\Big(-\frac{5\sqrt{34}}{34}\Big)^2-\Big(-\frac{3\sqrt{34}}{34}\Big)^2=\frac{9}{34}-\frac{25}{34}=-\frac{8}{17}$$