Question

:(tgx+ctgx)sin2x 2cos^2x/1+cos2x 2cos^2t-cos2t вычислить: 4sin15cos15 3sin^2a - 2sin(pi-a)+3cos^2 3a при a=pi/6

Answer 1

$(tgx+ctg x)\sin2x=( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} )2\sin x\cos x=2\sin^2x+2\cos^2x=2$

$\frac{2\cos^2x}{1+\cos2x} = \frac{2\cos^2x}{1+2\cos^2x-1} = \frac{2\cos^2x}{2\cos^2x} =1$

$2\cos^2t-\cos2t=2\cos^2x-2\cos^2x+1=1$

$4\sin15а\cos15а=2\sin30а=2\cdot \frac{1}{2} =1$

$3\sin^2 \alpha -2\sin( \pi - \alpha )+3\cos^23a=3\sin^2 \alpha -2\sin \alpha +3\cos^23 \alpha \\ \alpha = \frac{ \pi }{6} \\ 3\sin^2\frac{ \pi }{6}-2\sin\frac{ \pi }{6}+3\cos^2\frac{ \pi }{2}= \frac{3}{4} -1+0=- \frac{1}{4}$