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

Answer 2

Let's simplify the expression:

2sin²α - 1 + cos²α + (1 - sinα)(1 + sinα)

First, we can use the trigonometric identity: sin²α + cos²α = 1.

So, the expression becomes:

2sin²α - 1 + (1 - sinα)(1 + sinα)

Now, we can expand the last term:

2sin²α - 1 + (1 - sin²α)

Now, simplify further:

2sin²α - 1 + 1 - sin²α

Combine like terms:

(2sin²α - sin²α) - 1 + 1

sin²α - 1 + 1

Now, the 1 and -1 in the expression cancel out:

sin²α

So, the simplified expression is just sin²α.