Докажите, помогите, пожалуйста[latex]1-4sin^2x=4sin (\frac{pi}{6} - x ) sin( \frac{pi}{6}+x) 3-4cos^2y= -4sin( \frac{pi}{6} +y)sin( \frac{pi}{6} -y)[/latex]

Question

Докажите, помогите, пожалуйста[latex]1-4sin^2x=4sin (\frac{pi}{6} - x ) sin( \frac{pi}{6}+x) 3-4cos^2y= -4sin( \frac{pi}{6} +y)sin( \frac{pi}{6} -y)[/latex]

Докажите, помогите, пожалуйста [latex]1-4sin^2x=4sin (\frac{pi}{6} - x ) sin( \frac{pi}{6}+x) 3-4cos^2y= -4sin( \frac{pi}{6} +y)sin( \frac{pi}{6} -y)[/latex]

Гость

Ответ(ы) на вопрос:

Accepted Answer

[latex]1-4sin^2x=4sin(\frac{\pi}{6}-x)sin(\frac{\pi}{6}+x)\\\\ sin(\frac{\pi}{6}-x)sin(\frac{\pi}{6}+x) = \frac{cos2x-cos\frac{\pi}{3}}{2}=\frac{cos2x-0.5}{2}\\ 4*\frac{cos2x-0.5}{2}=2(cos2x-0.5)=2cos2x-1\\ 2(1-2sin^2a)-1=2-4sin^2a-1=1-4sin^2a[/latex] [latex]3-4cos^2y=-4sin(\frac{\pi}{6}+y)sin(\frac{\pi}{6}-y)\\ -4*\frac{cos2y-cos\frac{\pi}{3}}{2}=-2(cos2y-0.5)=-2(2cos^2y-1.5)=-4cos^2y+3[/latex]