Question

IF X =a cos theta +b sin theta and y =a sin theta -b cos theta ,proves that X square+ yswuare =asquare+b square

Answer 1

Given

$\sf x = \sf a cos\theta +bsin\theta$

$\sf y = asin\theta-bcos\theta$

To Prove

$\sf x^{2} + y^{2} = a^{2} +b^{2}$

Solving the LHS side, we get,

$\implies \sf x^{2} + y^{2}$

$\implies ( \sf a cos\theta +bsin\theta)^{2} + \sf (asin\theta-bcos\theta)^{2}$

$\implies \sf a^{2}cos^{2}\theta + 2abcos\theta\sin\theta + b^{2}sin^{2}\theta + a^{2}sin^{2}\theta - 2abcos\theta\sin\theta + b^{2}cos^{2}\theta$

$\implies \sf a^{2}cos^{2}\theta + b^{2}sin^{2}\theta + a^{2}sin^{2}\theta + b^{2}cos^{2}\theta$

$\implies \sf a^{2}cos^{2}\theta + a^{2}sin^{2}\theta + b^{2}sin^{2}\theta + b^{2}cos^{2}\theta$

$\implies \sf a^{2}(sin^{2}\theta+ cos^{2}\theta) + b^{2} (sin^{2}\theta+ cos^{2}\theta)$

$\implies \sf a^{2} +b^{2} (Since, \ sin^{2}\theta + cos^{2}\theta = 1)$

= RHS

                                                       

Answer 2

GIVEN :-

X = a cosθ + b sinθ

Y = a sinθ - b cosθ

TO PROVE :-

X² + Y² = a² + b²

PROOF :-

LHS :-

X² + Y²

= (a cosθ + b sinθ)² + (a sinθ - b cosθ)²

= (a²cos²θ + b²sin²θ + 2ab sinθcosθ) + (a²sin²θ + b²cos²θ - 2ab sinθcosθ)

= a²cos²θ + a²sin²θ + b²sin²θ + b²cos²θ + 2ab sinθcosθ - 2ab sinθcosθ

= a²(cos²θ + sin²θ) + b²(sin²θ + cos²θ)

= a²(1) + b²(1)

= a² + b² = RHS

$\rule{200}{2}$

Identities used :-

• (a + b)² = a² + b² + 2ab

• (a - b)² = a² + b² - 2ab

• sin²θ + cos²θ = 1