삼각함수 전체 문제

$\begin{array}{ccc}\frac{\sin (2\theta )}{1+\cos (2\theta )}{\tan }^{2}\theta & = & \frac{2\sin (\theta )\cos (\theta )}{1+{\cos }^{2}\theta -{\sin }^{2}\theta }{\tan }^{2}\theta = \\ \frac{2\sin (\theta )\cos (\theta )}{2{\cos }^{2}\theta }{\tan }^{2}\theta & = & \frac{\sin (\theta )}{\cos \,\theta }{\tan }^{2}\theta = \\ \tan (\theta ){\tan }^{2}\theta & = & {\tan }^{3}\,\theta \end{array}$

$\begin{array}{c}\frac{2\tan \,x}{1+{\tan }^{2}x}=\frac{\frac{2\sin \,x}{\cos \,x}}{1+\frac{{\sin }^{2}x}{{\cos }^{2}x}}=\frac{\frac{2\sin \,x}{\cos \,x}}{\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}}= \\ \frac{2\sin \,x}{\cos \,x}.\frac{{\cos }^{2}x}{1}=2\sin \,x\cos \,x=\sin (2x)\end{array}$

$\frac{2\sin \,x\cos \,x}{2{\cos }^{2}x-1}=\frac{\sin (2x)}{\cos (2x)}=\tan (2x)$

$\begin{array}{ccc}\sin (x+2x) & = & \sin \,x\cos (2x)+\sin (2x)\cos \,x \\ & = & \sin \,x({\cos }^{2}x-{\sin }^{2}x)+2\sin \,x\cos \,x\cos \,x \\ & = & \sin \,x{\cos }^{2}x-{\sin }^{3}x+2\sin \,x{\cos }^{2}x \\ & = & 3\sin \,x{\cos }^{2}x-{\sin }^{3}x\end{array}$

$\begin{array}{ccc}\frac{1+\cos (2t)}{\sin (2t)-\cos t} & = & \frac{1+2{\cos }^{2}t-1}{2\sin t\cos t-\cos t} \\ & = & \frac{2{\cos }^{2}t}{\cos t(2\sin t-1)} \\ & = & \frac{2\cos t}{2\sin t-1}\end{array}$

$\begin{array}{ccc}({\cos }^{2}(4x)-{\sin }^{2}(4x)-\sin (8x))({\cos }^{2}(4x)-{\sin }^{2}(4x)+\sin (8x)) & = & \\ & = & (\cos (8x)-\sin (8x))(\cos (8x)+\sin (8x)) \\ & = & {\cos }^{2}(8x)-{\sin }^{2}(8x) \\ & = & \cos (16x)\end{array}$

$$\begin{array}{c}2\,\sin (7x)-2\,\sin x=2\,\sin (4x+3x)-2\,\sin (4x-3x)= \\ 2(\sin (4x)\cos (3x)+\sin (3x)\cos (4x))-2(\sin (4x)\cos (3x)-\sin (3x)\cos (4x))= \\ 2\,\sin (4x)\cos (3x)+2\,\sin (3x)\cos (4x))-2\,\sin (4x)\cos (3x)+2\,\sin (3x)\cos (4x))= \\ 4\,\sin (3x)\cos (4x) \\ \end{array}$$

$\begin{array}{ccc}\sin \,x+\sin (3x) & = & 2\,\sin \left(\frac{4x}{2}\right)\cos \left(\frac{-2x}{2}\right)= \\ 2\,\sin (2x)\cos \,x & = & 2(2\,\sin \,x\,\cos \,x)\cos \,x= \\ 4\,\sin \,x\,{\cos }^{2}\,x & & \end{array}$

$\begin{array}{c}2\,\tan \,x\,\cos (3x)=\frac{2\,\sin \,x\,\cos (3x)}{\cos \,x}=\frac{2(.5(\sin (4x)-\sin (2x)))}{\cos \,x}= \\ \frac{1}{\cos \,x}(\sin (4x)-\sin (2x))=\sec \,x(\sin (4x)-\sin (2x))\end{array}$

Start with $\cos \,x+\cos \,y.$ Make a substitution and let $x=\alpha +\beta$ and let $y=\alpha -\beta ,$ so $\cos \,x+\cos \,y$ becomes $\begin{array}{c} \\ \cos (\alpha +\beta )+\cos (\alpha -\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta +\cos \alpha \cos \beta +\sin \alpha \sin \beta = \\ 2\cos \,\alpha \cos \,\beta \end{array}$

Since $x=\alpha +\beta$ and $y=\alpha -\beta ,$ we can solve for $\alpha$ and $\beta$ in terms of x and y and substitute in for $2\cos \alpha \cos \beta$ and get $2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right).$