# If sinϴ+cosϴ = √2 cosϴ prove that cosϴ-sinϴ = √2 sinϴ, (ϴ is an acute angle)

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Given `sin theta + cos theta = sqrt(2)cos theta` prove `cos theta - sin theta = sqrt(2)sin theta`

`sin theta + cos theta = sqrt(2)cos theta` square both sides

`sin^2theta + cos^2 theta + 2sintheta costheta=2cos^2theta`

`sin^2theta - cos^2theta + 2sinthetacostheta=0`

`-sin^2theta + cos^2theta -2sinthetacostheta=0` Add `2sin^2theta` to both sides

`sin^2theta+cos^2theta-2sinthetacostheta=2sin^2theta`

`(costheta-sintheta)^2=2sin^2theta`

`costheta-sintheta=sqrt(2)sintheta` as required.

Divide by `costheta` to get

` tantheta+1=sqrt(2)`

so

`tantheta=sqrt(2)-1`

`1/tantheta=cottheta`

Take the reciprocal of both sides

`cottheta=1/(sqrt(2)-1)`

`1/(sqrt(2)-1)*(sqrt(2)+1)/(sqrt(2)+1)=(sqrt(2)+1)/(2-1)=sqrt(2)+1`

So

` cottheta=sqrt(2)+1`

`costheta/sintheta=sqrt(2)+1`

`costheta/sintheta-1=sqrt(2)`

Multiply both sides by `sintheta`

`costheta-sintheta=sqrt(2)sintheta`

Which is what we were supposed to prove.

cosϴ +sinϴ = √2 cosϴ (given)

squaring on the both sides , we get

cos2ϴ + sin2ϴ + 2cosϴsinϴ = 2 cos2ϴ

sin2ϴ + 2cosϴsinϴ = cos2ϴ

cos2ϴ - 2cosϴsinϴ = sin2ϴ

cos2ϴ - 2cosϴsinϴ + sin2ϴ = 2 sin2ϴ (adding sin2ϴ on both sides)

(cosϴ - sinϴ)2 = 2 sin2ϴ

cosϴ-sinϴ = √2 sinϴ

Hence proved!!