Question

Sin^2a+sin^2b+ cos(a+b)*cos(a-b)=a. а-альфа в-бета

Answer 1

$\star \; \; cosx\cdot cosy=\frac{1}{2}(cos(x-y)+cos(x+y))\; \; \star \\\\\\sin^2 \alpha +sin^2 \beta +cos( \alpha + \beta )\cdot cos( \alpha - \beta )=\\\\=sin^2 \alpha +sin^2 \beta +\frac{1}{2}(cos2 \beta +cos2 \alpha )=\\\\=\quad \star\; \; [\; cos2x=1-2sin^2x\; ]\; \; \star =\\\\=sin^2 \alpha +sin^2 \beta +\frac{1}{2}(1-2sin^2 \beta +1-2sin^2 \alpha )=\\\\=sin^2 \alpha +sin^2 \beta +\frac{1}{2}-sin^2 \beta +\frac{1}{2}-sin^2 \alpha =1$