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Question 1

Question 2

Решите задачу:

$\sin \alpha \sin \beta = \frac{1}{2} (\cos( \alpha - \beta )-\cos( \alpha + \beta )) \\\ \sin(x+ \frac{ \pi }{8} )\sin(x- \frac{ \pi }{8} )= \frac{1}{2} (\cos( x+ \frac{ \pi }{8} -x+ \frac{ \pi }{8} )-\cos( x+ \frac{ \pi }{8}+x- \frac{ \pi }{8} ))= \\\ =\frac{1}{2} (\cos\frac{ \pi }{4}-\cos2 x)=\frac{1}{2} (\frac{\sqrt{2}}{2}-\cos2 x)=\frac{ \sqrt{2} }{4}-\frac{1}{2}\cos2 x \\\ \sin75\sin15= \frac{1}{2}(\cos( 75-15)-\cos(75+15))= \\\ =\frac{1}{2} (\cos60-\cos90)=\frac{1}{2}(\frac{1}{2} -0)= \frac{1}{4}$

$\cos \alpha \cos \beta = \frac{1}{2} (\cos( \alpha + \beta )+\cos( \alpha - \beta )) \\\ \cos(x+y) \cos (x-y)= \frac{1}{2} (\cos( x+y+x-y)+\cos( x+y-x+y ))= \\\ =\frac{1}{2} (\cos2x+\cos 2y)=\frac{1}{2} \cos2x+\frac{1}{2}\cos 2y \\\ \cos40\cos20= \frac{1}{2} (\cos( 40+ 20)+\cos( 40- 20))= \\\ =\frac{1}{2} (\cos60+\cos20)=\frac{1}{2} (\frac{1}{2}+\cos20)=\frac{1}{4}+\frac{1}{2}\cos20$

$\sin\alpha\cos\beta=\frac{1}{2}(\sin( \alpha+\beta)+\sin(\alpha-\beta)) \\\ \sin (30+x)\cos (30-x)= \frac{1}{2} (\sin( 30+x+30-x )+\sin( 30+x- \\\ -30+x))=\frac{1}{2}(\sin60+\sin2x)=\frac{1}{2}(\frac{\sqrt{3}}{2}+\sin2x)=\frac{\sqrt{3}}{4}+\frac{1}{2}\sin2x \\\ \sin(\frac{\pi}{4}-y)\cos(\frac{\pi}{4}+y)=\frac{1}{2}(\sin(\frac{\pi}{4}-y+\frac{\pi}{4}+y )+\sin(\frac{\pi}{4}-y-\frac{ \pi}{4}-y))= \\\ =\frac{1}{2}(\sin\frac{\pi}{2}-\sin2y)=\frac{1}{2}(1-\sin2y)=\frac{1}{2}-\frac{1}{2}\sin2y$

Question 3

спасибо!