Question

Решить,) дано: а)sin60*cos135*tg120 б)cos60-2sin^2 135+cos^2 150 эти выражения.

Answer 1

$\sin60^0* \cos135^0*\tan120^0=\frac{\sqrt{3}}{2} \cos135^0*\tan120^0$

$\frac{\sqrt{3}}{2} \cos135^0*\tan120^0=\frac{\sqrt{3}}{2} \cos(90^0+45^0)*\tan120^0$

$\frac{\sqrt{3}}{2} \cos(90^0+45^0)*\tan120^0=$

$=\frac{\sqrt{3}}{2}( \cos90^0\cos45^0-\sin90^0*\sin45^0)*\tan120^0$

$\frac{\sqrt{3}}{2}\left(-\frac{\sqrt{2}}{2}\right)*\tan120^0=-\frac{\sqrt{6}}{4}*\tan120^0$

$-\frac{\sqrt{6}}{4}*\tan120^0=-\frac{\sqrt{6}}{4}*\frac{\sin120^0}{\cos 120^0}$

$-\frac{\sqrt{6}}{4}*\frac{\sin(90^0+30^0)}{\cos (90^0+30^0)}=$

$=-\frac{\sqrt{6}}{4}*\frac{\sin90^0*\cos 30^0+\cos90^0\sin30^0}{\cos (90^0+30^0)}$

$-\frac{\sqrt{6}}{4}*\frac{\sin90^0*\cos 30^0+\cos90^0\sin30^0}{\cos (90^0+30^0)}=-\frac{\sqrt{6}}{4}*\frac{\frac{\sqrt{3}}{2}}{\cos (90^0+30^0)}$

$-\frac{3\sqrt{2}}{8}*\frac{1}{\cos (90^0+30^0)}=-\frac{3\sqrt{2}}{8}*\frac{1}{\cos 90^0\cos30^0-\sin90^0\sin30^0}$

$-\frac{3\sqrt{2}}{8}*\frac{1}{\cos 90^0\cos30^0-\sin90^0\sin30^0}=-\frac{3\sqrt{2}}{8}*\frac{1}{-\frac{1}{2}}$

$-\frac{3\sqrt{2}}{8}*\frac{1}{-\frac{1}{2}}=2*\frac{3\sqrt{2}}{8}$

$2*\frac{3\sqrt{2}}{8}=\frac{3\sqrt{2}}{4}$

$\cos60^0-2\sin^2 135^0+\cos^2 150^0=\frac{1}{2}-2\sin^2 135^0+\cos^2 150^0$

По формуле двойного угла

$\frac{1}{2}-2\sin^2 135^0+\cos^2 150^0=\frac{1}{2}-(1-\cos(2*135^0))+\cos^2150^0$

$\frac{1}{2}-(1-\cos(2*135^0))+\cos^2150^0=\frac{1}{2}-1+\cos 270^0+\cos^2150^0$

$\frac{1}{2}-1+\cos 270^0+\cos^2150^0=-\frac{1}{2}+\cos 270^0+\cos^2150^0$

Снова по формуле двойного угла

$-\frac{1}{2}+\cos 270^0+\cos^2150^0=-\frac{1}{2}+\cos 270^0+\frac{1}{2}(1+\cos(2*150^0))$

$-\frac{1}{2}+\cos 270^0+\frac{1}{2}(1+\cos(2*150^0))=\cos 270^0+\frac{1}{2}\cos300^0$

$\cos 270^0+\frac{1}{2}\cos300^0=0+\frac{1}{2}\cos300^0$

$\frac{1}{2}\cos300^0=\frac{1}{2}\cos(360^0-60^0)$

$\frac{1}{2}\cos(360^0-60^0)=\frac{1}{2}(\cos 360^0\cos 60^0+\sin 360^0\sin60^0)=\frac{1}{4}$