Question

sin theta=a-b/a+b,cos theta=?

Answer 1

Given, sinθ = $\dfrac{a-b}{a+b}$

Square on both sides

⇒ sin²θ = $\dfrac{(a-b)^{2}}{(a+b)^{2}}$

sin²A = 1 - cos²A     [ From trigonometric identities ]

⇒ 1 - cos²θ = $\dfrac{(a-b)^2}{(a+b)^2}$

⇒ 1 - $\dfrac{(a-b)^2}{(a+b)^2}$= cos²θ

⇒ $\dfrac{1( a +b)^2 - ( a-b)^2}{(a+b)^2}$ = cos²θ

⇒ $\dfrac{( a +b)^2 - ( a-b)^2}{(a+b)^2}$ = cos²θ

( a + b )^2 = a^2 + b^2 + 2ab  [ From expansion ]

( a - b )^2 = a^2 + b^2 + 2ab   [ From expansion ]

$\Rightarrow \dfrac{a^2 + b^2 + 2ab - ( a^2 + b^2 - 2ab) }{ (a+b)^2} = cos^2 \theta$

$\Rightarrow \dfrac{a^2 + b^2 + 2ab - a^2 - b^2 + 2ab) }{ (a+b)^2} = cos^2 \theta$

⇒ $\dfrac{4ab}{(a+b)^2}$  = cos²θ

⇒ $\dfrac{4ab}{(a+b)^2} =$ cos²θ

Square root on both sides

⇒ $\sqrt{\dfrac{4ab}{(a+b)^2} }= \sqrt{cos^2 \theta}$

⇒ $\dfrac{2\sqrt{ab}}{a+b}$ = cosθ

Therefore the value of cosθ satisfying the given equation is $\dfrac{2\sqrt{ab}}{a+b}$

Answer 2

cos theta=root 1-sinsq.theta
=root 1-(a-b)sq./(a+b)sq.
=root (a+b)sq.-(a-b)sq./(a+b)sq.
=root a^2+b^2+2ab-a^2-b^2+2ab /(a+b)sq.
=root 4ab/(a+b)sq.
=(2/a+b)root ab