Question

Prove that if x = a sin theta+ b cos theta
and
y = a cos theta-b sin theta
, then x2 + y2 = a2+ b2​

Answer 1

Solution

Given

$\sf \: x = a \sin( \alpha ) + b \cos( \alpha )$

$\sf \: y = a \cos( \alpha ) - b \sin( \alpha )$

To Prove

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

LHS

$\sf \: x {}^{2} + {y}^{2} \\ \\ \dashrightarrow \: \sf \: \bigg(a \sin( \alpha ) + b \cos( \alpha ) \bigg) {}^{2} + \bigg(a \cos( \alpha ) - b \sin( \alpha ) \bigg) {}^{2} \\ \\ \dashrightarrow \: \sf \: \bigg( {a}^{2} {sin}^{2} \alpha + {b}^{2} {cos}^{2} \alpha + 2ab \: sin( \alpha )cos( \alpha ) \bigg) + \bigg( {a}^{2} {cos}^{2} \alpha + {b}^{2} {sin}^{2} \alpha - 2ab \: sin( \alpha ) \: cos( \alpha ) \bigg) \\ \\ \dashrightarrow \sf \: {a}^{2} ( {sin}^{2} \alpha + cos {}^{2} \alpha ) + {b}^{2} ( {sin}^{2} \alpha + {cos}^{2} \alpha ) \\ \\ \dashrightarrow \: \sf \: {a}^{2} + {b}^{2}$

Hence,Proved

NoTE

Trigonometric Identities

• sin²∅ + cos∅ = 1

• sec²∅ - tan²∅ = 1

• cosec²∅ - cot²∅ = 1

Algebraic Identities

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

Answer 2

Check the attachment.