Question

If x = a cos theta + b sin theta and y = a sin theta - b cos theta , prove that x^2 +y^2 = a^2+ b^2.

Answer 1

hey here is your answer.

Answer 2

Your Answer:

Given:-

• x = a cos Θ + b sin Θ
• y = a sin Θ - b cos Θ

To Prove:-

• x² +y² = a²+ b²

Solution:-

Solving LHS

$\tt x^{2}+y^{2}\\\\ \tt = (a \cos \theta + b \sin \theta)^{2}+ (a \sin \theta - b \cos \theta)^{2}\\\\ \tt =(a \cos\theta)^{2}+(b\sin\theta)^{2}+2ab\sin\theta\cos\theta+(a \cos\theta)^{2}+(b\sin\theta)^{2}-2ab\sin\theta\cos\theta\\\\ \tt =a^{2}{\cos}^{2}\theta+b^{2}{\sin}^{2}\theta+a^{2}{\cos}^{2}\theta+b^{2}{\sin}^{2}\theta\\\\ \tt = a^{2}({\cos}^{2}\theta+{\sin}^{2}\theta)+b^{2}({\cos}^{2}\theta+{\sin}^{2}\theta)\\\\ = \tt a^{2}+b^{2}\:\:\:\:\:\:\:\:\:\:(\because {\cos}^{2}\theta+{\sin}^{2}\theta=1)$

LHS=RHS

proved