Sqrt 32-sqrt 128*sin^2 (9π/8)

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вопрос

sqrt(32) - sqrt(128)*sin^2(9π/8)

First, simplify the square roots:

sqrt(32) = sqrt(162) = sqrt(16)sqrt(2) = 4*sqrt(2)

sqrt(128) = sqrt(642) = sqrt(64)sqrt(2) = 8*sqrt(2)

Now substitute these values back into the original expression:

4sqrt(2) - 8sqrt(2)*sin^2(9π/8)

Next, note that sin^2(π - x) = sin^2(x). Therefore, sin^2(9π/8) = sin^2(π + π/8) = sin^2(π/8).

Substitute this back into the expression:

4sqrt(2) - 8sqrt(2)*sin^2(π/8)

Since sin(π/8) = sin(22.5 degrees) = sqrt(2-sqrt(2))/2, the final expression becomes:

4sqrt(2) - 8sqrt(2)*((2-sqrt(2))/2)

Simplify this expression:

4sqrt(2) - 4sqrt(2)*(2 - sqrt(2))

= 4sqrt(2) - 8sqrt(2) + 4*sqrt(2)

= 4sqrt(2) + 4sqrt(2) - 8*sqrt(2)

= 8sqrt(2) - 8sqrt(2)

= 0

Therefore, Sqrt 32-sqrt 128*sin^2(9π/8) simplifies to 0.