Naloge

$\sin\alpha\cdot\sin\beta\cdot\sin\gamma=$

$=\sin\alpha\cdot\sin\beta\cdot\sin(180°-(\alpha+\beta))=$

$=\sin\alpha\cdot\sin\beta\cdot\sin(\alpha+\beta)=$

$=-\frac{1}{2}\big (\cos(\alpha+\beta)-\cos(\alpha-\beta)\big)\cdot\sin(\alpha+\beta)=$

$=-\frac{1}{2}\big (\cos(\alpha+\beta)\sin(\alpha+\beta)-\cos(\alpha-\beta)\sin(\alpha+\beta))=$

Oba zmnožka razčlenimo in dobimo

$=\frac{1}{4}\big (\sin(2\alpha)+\sin(2\beta)-\sin(2\alpha+2\beta))$

$\displaystyle \tan52,5°\cdot\tan7,5°=\frac{\sin52,5°\cdot\sin7,5°}{\cos52,5°\cdot\cos7,5°}=$

$\displaystyle \qquad =\frac{-(\cos60°-\cos45°)}{\cos60°+\cos45}°=\frac{1-\sqrt{2}}{1+\sqrt{2}}=3-2\sqrt{2}$

$\displaystyle \tan\frac{\pi}{12}\cdot\tan\frac{5\pi}{12}=\frac{\sin\frac{\pi}{12}}{\cos\frac{\pi}{12}}\cdot\frac{\sin\frac{5\pi}{12}}{\cos\frac{5\pi}{12}}=$

$\displaystyle \quad =\frac{-(\cos\frac{6\pi}{12}-\cos\frac{4\pi}{12})}{\cos\frac{6\pi}{12}+\cos\frac{4\pi}{12}}=\frac{-(\cos\frac{\pi}{2}-\cos\frac{\pi}{3})}{\cos\frac{\pi}{2}+\cos\frac{\pi}{3}}=1$

$\sin(6x+20°)+\sin(2x)$

$4\cos(x+\frac{\pi}{2})\cdot\sin(x-\frac{\pi}{2})=$

$\quad =2\cdot(\sin(2x)+\sin(-\pi))=2\sin(2x)$

$2\cos x\cdot\cos(x+\frac{\pi}{4})+2\sin x\cdot\sin(x+\frac{\pi}{4})=\ldots$

$\quad = 2\cos(-\frac{\pi}{4})=\sqrt{2}$