Question

If sin θ +2cos θ =1, then prove that 2 sin θ -cos θ =2.

Answer 1

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.

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

Answer 2

hence proved.................