Question

Let cosθ+sinθ = √2 cosθ , then prove that cosθ - sinθ = √2 sinθ

Answer 1

Answer:

Step-by-step explanation:

Let θ = x

cos x  + sin x =  √2 cos x

squaring on both side, we get......

cos2x + sin2x + 2cosxsinx = 2cos2x

2sinxcosx = 2cos2x - cos2x - sin2x

2sinxcosx = cos2x - sin2x

2sinxcosx = (cosx+sinx) (cosx - sinx)

2sinxcosx = (root2 cosx) (cosx - sinx)

2sinxcosx/root2 cosx = cosx - sinx

√2 sinx = cosx - sinx

Answer 2

$\underline{\underline{\bold{Question:}}}$

$\bold{If\:cos\theta+sin\theta=\sqrt{2}\:cos\theta }\\\\\bold{then\:prove\:that\quad\:cos\theta-sin\theta=\sqrt2\:sin\theta.}$

$\bold{\underline{Solution:}}$

$\bold{Given:}\\\\\\\bold{cos\theta+sin\theta=\sqrt2\:cos\theta}\\\\\\\implies{\bold{sin\theta=\sqrt2\:cos\theta-cos\theta}}\\\\\\\implies{\bold{sin\theta=cos\theta(\sqrt2-1)}}\\\\\\\underline{\bold{Multiply\:by\:(\sqrt2+1)\:both\:side\:,}}\\\\\\\implies{\bold{sin\theta(\sqrt2+1)=cos\theta(\sqrt2-1)(\sqrt2+1)}}\\\\\\\boxed{\bold{a^2-b^2=(a+b)(a-b)}}\\\\\\\implies{\bold{\sqrt2\:sin\theta+sin\theta=cos\theta}}\\\\\\\implies{\bold{\sqrt2\:sin\theta=cos\theta-sin\theta}}$

$\bold{Proved.}\\\\\\\boxed{\boxed{\bold{cos\theta-sin\theta=\sqrt2\:sin\theta.}}}$