Cos^2(-x)+sin(-x)=2-sin^2x

Предыдущий
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вопрос

To prove this identity, we can first use the even and odd properties of cosine and sine functions:

cos(-x) = cos(x) (even
sin(-x) = -sin(x) (odd)

Therefore, the given expression can be written as:

cos^2(-x) + sin(-x) = cos^2(x) - sin(x)

Next, we can use the Pythagorean identity cos^2(x) + sin^2(x) = 1 to simplify the expression further:

cos^2(x) - sin(x
= cos^2(x) - (1 - cos^2(x)) [Using Pythagorean identity sin^2(x)=1-cos^2(x)
= cos^2(x) - 1 + cos^2(x
= 2cos^2(x) - 1

Therefore, the given expression cos^2(-x) + sin(-x) simplifies to 2cos^2(x) - 1.

Now, we can use the Pythagorean identity again to simplify the right side of the equation:

2 - sin^2(x
= 2 - (1 - cos^2(x)
= 2 - 1 + cos^2(x
= 1 + cos^2(x)

So, the right side simplifies to 1 + cos^2(x).

Since we have shown that cos^2(-x) + sin(-x) simplifies to 2cos^2(x) - 1, we can rewrite the original expression as:

2cos^2(x) - 1 = 1 + cos^2(x)

Simplifying this expression further, we get:

2cos^2(x) - 1 = 1 + cos^2(x
cos^2(x) = 2

Therefore, the identity cos^2(-x) + sin(-x) = 2 - sin^2(x) is proved to be true.