Question

15. 2sin²α-1+cos²α+(1-sinα)(1+sinα)​

Answer 1

Решение:

$2sin^{2}\alpha-1+cos^{2}\alpha+(1-sin\alpha)(1+sin\alpha)= sin^{2}\alpha+sin^{2}\alpha-1+cos^{2}\alpha+((1)^{2}-(-sin\alpha)^{2})=sin^{2}\alpha+sin^{2}\alpha-1+cos^{2}\alpha+(1-sin^{2}\alpha)=sin^{2}\alpha+sin^{2}\alpha-1+cos^{2}\alpha+1-sin^{2}\alpha=1+sin^{2}\alpha-1+1-sin^{2}\alpha=1$

Ответ: 1