Question

Известно, что sin(a+b)=0.2 и cos(a-b)=0.3. Вычислите sin(a+45°)*sin(b+45°).

Answer 1

$\displaystyle sin(a+\pi /4)*sin(b+\pi /4)=\\\\=(sina*cos\pi /4+cosa*sin\pi /4)*(sinb*cos\pi /4+cosb*sin\pi /4)=\\\\=(\frac{\sqrt{2}}{2}(sina+cosa))*(\frac{\sqrt{2}}{2}(sinb+cosb))=\\\\= \frac{1}{2} (sina*sinb+cosa*sinb+sina*cosb+cosa*cosb)= \\\\=\frac{1}{2}(sina*cosb+cosa*sinb+cosa*cosb+sina*sinb)=\\\\=\frac{1}{2}(sin(a+b)+cos(a-b))=\frac{1}{2}(0.2+0.3)= 0.25$