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Find sin^2(pi/8) - cos^4(3pi/8)

User Dan Walker
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1 Answer

5 votes

Recall the following identities:

sin²(x) = (1 - cos(2x))/2

cos²(x) = (1 + cos(2x))/2

Then

sin²(π/8) = (1 - cos(π/4))/2 = (1 - 1/√2)/2

cos⁴(3π/8) = (cos²(3π/8))² = ((1 + cos(3π/4))/2)² = ((1 - 1/√2)/2)²

and so

sin²(π/8) - cos⁴(3π/8) = (1 - 1/√2)/2 - ((1 - 1/√2)/2)²

… = (1 - 1/√2)/2 • (1 - (1 - 1/√2)/2)

… = (2 - √2)/4 • (1 - (2 - √2)/4)

… = (2 - √2)/16 • (4 - 2 + √2)

… = (2 - √2)(2 + √2)/16

… = (4 - 2)/16

… = 1/8

User Bosticko
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