Question

If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.

Answer 1

(cosθ + sinθ)² = (`sqrt2`. cosθ)²

cos²θ +  sin²θ + 2.cosθ.sinθ = 2cos²θ

1 + 2.cosθ.sinθ = 2cos²θ

2.cosθ.sinθ = 2cos²θ − 1

(cosθ.sinθ)² = cos²θ + sin²θ − 2.cosθ.sinθ

= 1 − (2.cos²θ − 1)

= 1 − 2.cos²θ +1

= 2 − 2.cos²θ

= 2(1 − cos²θ)

cosθ − sinθ = `sqrt(2sin^2θ)`

= `sqrt2`sinθ

Hence proved

Answer 2

Given:

cosθ + sinθ = `sqrt2`cosθ

squaring on both sides, we get,

cos²θ + sin²θ + 2sinθ cosθ = 2cos²θ

cos²θ − sin²θ = 2cosθ sinθ

(cosθ + sinθ)(cosθ − sinθ) = 2cosθ sinθ

`sqrt2`cosθ (cosθ − sinθ) = 2cosθ sinθ   ...[ Given cosθ + sinθ = `sqrt2`cosθ]

∴cosθ − sinθ = `sqrt2`sinθ

Hence proved