# Prove that sin(5π/24) = [√(6+3√2)-√(2-√2)]/4

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### 2 Answers

You may write `5pi/24 = pi/24 + 4pi/24 =gt 5pi/24 = pi/24 + pi/6`

You should also consider `pi/24 = (pi/12)/2` .

Hence, `sin 5pi/24 = sin (pi/24 + pi/6) `

Expanding the sine of the sum yields:

`sin 5pi/24 = sin pi/24*cos pi/6 + sin pi/6*cos pi/24`

You need to evaluate sin `pi/24` such that:

`sin pi/24 = sqrt((1 - cos(pi/12))/2)`

You need to evaluate `cos pi/24` such that:

`cos pi/24 = sqrt((1 + cos(pi/12))/2)`

You need to consider `pi/12 = (pi/6)/2` , hence, you should evaluate `cos(pi/12)` such that:

`cos(pi/12) = sqrt((1 + cos(pi/6))/2)`

`cos(pi/12) = sqrt((1 + sqrt3/2)/2) =gt cos(pi/12) = (2+sqrt3)/2`

Substituting `(2+sqrt3)/2` for `cos(pi/12)` yields:

`cos pi/24 = sqrt((1 + (2+sqrt3)/2)/2)`

`cos pi/24 = (sqrt(4+sqrt3))/2`

`sin pi/24 = (sqrt(4-sqrt3))/2`

Substituting `(sqrt(4-sqrt3))/2` for `sin pi/24 ` and `(sqrt(4+sqrt3))/2` for cos `pi/24` such that:

`sin 5pi/24 =sqrt3(sqrt(4-sqrt3))/4 + (sqrt(4+sqrt3))/4`

`sin 5pi/24 = (sqrt3(sqrt(4-sqrt3) + sqrt(4+sqrt3)))/4`

Hence, evaluating the value of `sin 5pi/24` yields `sin 5pi/24 = (sqrt3(sqrt(4-sqrt3) + sqrt(4+sqrt3))/4.`

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Got the answer...

L:H:S = sin(5π/24)

= sin(π/3 - π/8)

= sin π/3.cos π/8 - cos π/3.sin π/8

**➯ cos 2θ = 2cos²θ-1 ∴ cos θ = √ [(cos2θ+1)/2]**

**➯ cos 2θ = 1-2sin²θ ∴ sin θ = √ [(1-cos2θ)/2]**

= sin π/3.√ [(cos π/4+1)/2] - cos π/3. √ [(1-cos π/4)/2]

= (√3/2) * √ [(1+√2)÷ 2√2] - (1/2) * √ [(√2-1)÷ 2√2]

= (√3/2) * √ [(√2+2)÷ 4] - (1/2) * √ [(2-√2)÷ 4]

= (√3/4) * √ (√2+2) - (1/4) * √ (2-√2)

= (1/4) * √ (3√2+6) - (1/4) * √ (2-√2)

= [ √ (3√2+6) - √ (2-√2) ] / 4

= **[ √(6+3√2) - √(2-√2) ] / 4**

= R:H:S