Question

Помогите доказать тождество

Answer 1

Ответ:

Пошаговое объяснение:

sinxcosy=0,5(sin(x-y)+sin(x+y))

$4cos(\frac{\pi }{6}-a )sin(\frac{\pi }{3}-a )=4*0,5(sin((\frac{\pi }{3}-a )-(\frac{\pi }{6}-a ))+sin((\frac{\pi }{3}-a )+(\frac{\pi }{6}-a )))=2(sin\frac{\pi }{6}+sin(\frac{\pi }{2}-2a ))=2(0,5+cos2a)=1+2cos2a$

sin(x+y)=sinxcosy+sinycosx

sin3a=sin(a+2a)=sinacos2a+sin2acosa=sinacos2a+2sinacosacosa=sina(cos2a+2cos²a)=sina(cos2a+1+2cos²a-1)=sina(cos2a+1+cos2a)=sina(1+2cos2a)

sin3a/sina=sina(1+2cos2a)/sina=1+2cos2a

ч.т.д

Answer 2

Ответ:

$\star \ 4cos\Big(\dfrac{\pi}{6}-\alpha \Big)\cdot sin\Big(\dfrac{\pi}{3}-\alpha \Big)=4\cdot \dfrac{1}{2}\cdot \Big(sin\Big(\dfrac{\pi}{6}-\alpha +\dfrac{\pi}{3}-\alpha \Big)+sin\Big(\dfrac{\pi}{3}-\alpha -\dfrac{\pi}{6}+\alpha \Big)\Big)=\\\\\\=2\cdot \Big(sin\Big(\dfrac{\pi}{2}-2\alpha \Big)+sin\dfrac{\pi}{6}\Big)=2\cdot \Big(cos2\alpha +\dfrac{1}{2}\Big)=2\, cos2\alpha +1=$

$=2(\underbrace{cos^2\alpha -sin^2\alpha }_{cos2\alpha })+(\underbrace{sin^2\alpha +cos^2\alpha }_{1})=2cos^2\alpha -2sin^2\alpha +sin^2\alpha +cos^2\alpha=$

$=3cos^2\alpha -sin^2\alpha =3(1-sin^2\alpha )-sin^2\alpha =3-3sin^2\alpha -sin^2\alpha =\\\\\\=3-4sin^2\alpha \ \ ;$

$\star \ \ \displaystyle \frac{sin3\alpha }{sin\alpha }=\frac{3sin\alpha -4sin^3\alpha }{sin\alpha }=\dfrac{sin\alpha \cdot (3-4sin^2\alpha )}{sin\alpha }=3-4sin^2\alpha \ ;\\\\\\\star \ \ 3-4sin^2\alpha =3-4sin^2\alpha$