Question

6 задача помогите решить

Answer 1

Ответ:

13/16

Объяснение:

решение на фотографии

Answer 2

$Sin^{6}\alpha+Cos^{6}\alpha=(Sin^{2}\alpha)^{3}+(Cos^{2}\alpha)^{3}=(Sin^{2}\alpha+Cos^{2}\alpha)(Sin^{4}\alpha-Sin^{2}\alpha Cos^{2} \alpha+Cos^{6}\alpha)=Sin^{4}\alpha-Sin^{2} \alpha Cos^{2}\alpha+Cos^{4}\alpha=(Sin^{2}\alpha+Cos^{2}\alpha)^{2}-3Sin^{2}\alpha Cos^{2}\alpha=1-3Sin^{2}\alpha Cos^{2}\alpha$

$(Sin\alpha+Cos\alpha )^{2}=(\frac{\sqrt{3} }{\sqrt{2}})^{2}\\\\Sin^{2}\alpha+2Sin\alpha Cos\alpha+Cos^{2}\alpha=\frac{3}{2}\\\\1+2Sin\alpha Cos\alpha=\frac{3}{2}\\\\2Sin\alpha Cos\alpha=\frac{1}{2}\\\\Sin\alpha Cos\alpha=\frac{1}{4}$

$1-3Sin^{2}\alpha Cos^{2}\alpha=1-3*(\frac{1}{4})^{2} =1-\frac{3}{16}=\frac{13}{16}$