# if `sin x +cos x=sqrt2*sinx` then prove `cosx-sinx=sqrt2*cosx`

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You need to square the identity `sin x + cos x = sqrt2*sin x` such that:

`(sin x + cos x)^2 = (sqrt2*sin x)^2`

`sin^2 x + 2sin x*cos x + cos^2 x = 2sin^2 x`

You need to use the basic formula of trigonometry such that:

`sin^2 x + cos^2 x = 1`

`1 + 2sin x*cos x = 2sin^2 x => sin 2x = 2sin^2 x- 1 => sin 2x = -cos 2x`

You need to prove that `cos x - sin x = sqrt2*cos x` , hence, squaring both sides yields:

`cos^2 x - 2sin x*cos x + sin^2 x = 2cos^2 x`

`1 - 2sin x*cos x = 2cos^2 x => -2sin x*cos x = 2cos^2 x - 1`

`-sin 2x = cos 2x => sin 2x = -cos 2x`

**Notice that the last line proves that using the condition `sin x + cos x = sqrt2*sin x` yields that `cos x - sin x = sqrt2*cos x` .**

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