# How to solve sin^2(a-45)-cos^2(a-45)+sin 2a ?The answer is 0 .

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### 1 Answer

You need to substitute the difference of squares `sin^2(a-45^o)-cos^2(a-45^o)` by the special product such that:

`sin^2(a-45^o)-cos^2(a-45^o) = (sin(a-45^o) - cos(a-45^o))(sin(a-45^o) + cos(a-45^o))`

You need to expand `sin(a-45^o) ` and `cos(a-45^o)` such that:

`sin(a-45^o) = sin a*cos 45^o - sin 45^o*cos a`

`sin(a-45^o) = sin a *sqrt2/2 - cos a*sqrt2/2`

`cos(a-45^o) = cos a*sqrt2/2 + sin a*sqrt2/2`

`sin(a-45^o) - cos(a-45^o) = sin a *sqrt2/2 - cos a*sqrt2/2 - cos a*sqrt2/2 -sin a*sqrt2/2`

`sin(a-45^o) - cos(a-45^o) = -2cos a*sqrt2/2`

`sin(a-45^o) - cos(a-45^o) = -cos a*sqrt2`

`sin(a-45^o)+ cos(a-45^o) = sin a *sqrt2/2 - cos a*sqrt2/2+ cos a*sqrt2/2+ sin a*sqrt2/2`

`sin(a-45^o) + cos(a-45^o) = sin a *sqrt2`

`sin^2(a-45^o)-cos^2(a-45^o) = -(cos a*sqrt2)(sin a *sqrt2) + sin 2a`

`sin^2(a-45^o)-cos^2(a-45^o) = -2 cos a*sin a + sin 2a`

You should notice that `2 cos a*sin a` is the expansion of `sin 2a` such that:

`sin^2(a-45^o)-cos^2(a-45^o) = - sin 2a + sin 2a`

`sin^2(a-45^o)-cos^2(a-45^o) = 0`

**Hence, evaluating `sin^2(a-45^o)-cos^2(a-45^o) + sin 2a => sin^2(a-45^o)-cos^2(a-45^o) + sin 2a = 0` .**

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