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Sum
Prove the following:
`sin pi^"c"/8 = 1/2sqrt(2 - sqrt(2))`
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Solution
We know that cos `pi/4 = 1/sqrt(2)`
Also `pi/8` lies in the first quadrant, hence sin `pi/8` is positive.
Now, cos 2θ = 1 − 2sin2 θ
By putting θ = `pi/8`, we get,
`cos pi/4 = 1 - 2sin^2 pi/8`
∴ `2sin^2 pi/8 = 1 - cos pi/4`
= `1 - 1/sqrt(2)`
= `(sqrt2-1)/sqrt2`
∴ `sin^2 pi/8 = (sqrt(2) - 1)/(2sqrt(2))`
= `(sqrt(2)(sqrt(2) - 1))/4`
= `(2 - sqrt(2))/4`
∴ `sin pi/8 = 1/2 sqrt(2-sqrt2)` ......`[∵ sin pi/8 "is positive"]`
Concept: Trigonometric Functions of Double Angles
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