Prove that:sin^2 12+sin^2 21+sin^2 39+sin^2 48=1+sin^2 9+sin^2 18All the angles are in degree measure.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember and use the half angle formula, `sin^2 alpha = (1 - cos 2 alpha)/2`  ,such that:

`sin^212^o = (1 - cos 24^o)/2 `

`sin^2 21^o = (1 - cos 42^o)/2 `

`sin^2 39^o = (1 - cos 78^o)/2 `

`sin^2 48^o = (1 - cos 96^o)/2 `

`sin^2 9^o = (1 - cos 18^o)/2 `

`sin^2 18^o = (1 - cos 36^o)/2 `

`1 - cos 24^o + 1 - cos 42^o +1 - cos 78^o + 1 - cos 96^o = 2 + 1 - cos 18^o + 1 - cos 36^o`

Reducing like terms yields:

`cos 24^o + cos 42^o + cos 78^o + cos 96^o = cos 18^o + cos 36^o`

You need to combine the terms such that:

`(cos 24^o + cos 96^o) + (cos 42^o + cos 78^o) = (cos 18^o + cos 36^o)`

You need to convert the sum into product such that:

`cos 24^o + cos 96^o = 2cos((24^o + 96^o)/2)*cos((24^o- 96^o)/2)`

`cos 24^o + cos 96^o = 2cos 60^o*cos(-36^o)`

You need to remember that the cosine function is even, hence `cos(-36^o)= cos(36^o)`

`cos 24^o + cos 96^o = 2cos 60^o*cos(36^o)`

You need to convert the sum into product such that:

`cos 42^o + cos 78^o = 2cos((42^o + 78^o)/2)*cos((42^o - 78^o)/2)`

`cos 42^o + cos 78^o = 2cos 60^o*cos(-18^o) ` `cos 42^o + cos 78^o = 2cos 60^o*cos18^o`

`cos 18^o + cos 36^o = 2cos((18^o + 36^o)/2)*cos((18^o - 36^o)/2)`

Substituting the products into the sum above yields:

`2cos 60^o*cos(36^o) + 2cos 60^o*cos18^o = 2cos 27^o*cos 9^o`

`2cos 60^o*(cos 18^o + cos 36^o) = cos 18^o + cos 36^o `

Reducing `cos 18^o + cos 36^o`  both sides yields:

`2cos 60^o = 1 => cos 60^o = 1/2 => 1/2 = 1/2`

Hence, the last line proves that the given expression is an identity.

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