Prove that sin pi/8=square root(2-square root2)/2.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll apply half angle identity:

sin (a/2) = sqrt [(1-cosa)/2]

But sin (pi/8) = sin [(pi/4)/2] = sqrt {[1 - cos (pi/4)]/2}

cos pi/4 = sqrt2/2

sin (pi/8) = sqrt {[1 - (sqrt2/2)]/2}

sin (pi/8) = sqrt [(2-sqrt2)/4]

sin (pi/8) = sqrt (2-sqrt2)/sqrt 4

sin (pi/8) = [sqrt (2-sqrt2)]/2

We notice that the given identity is verified, such as sin (pi/8) = [sqrt (2-sqrt2)]/2.

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