Question

Prove that (a sin theta -b cos theta)^2+ (a cos theta+ b sin theta )^2 = a^2+b^2

Answer 1

Let theta = x.

LHS

= (asinx - bcosx)² + ( acosx + b sinx)²

= a²sin²x+ b²cos²x-2absinx(cosx) + a²cos²x + b² sin²x + 2absinx(cosx)

= a²(sin²x+cos²x) + b²(sin²x + cos²x)

= a² (1 ) + b² ( 1)

= a² + b²

= RHS.


Formulae : (a+b)² = a²+b²+2ab
(sin²x+cos²x) = 1

Answer 2

To prove ---

( a sin ∅ - b cos ∅)²+ ( a cos ∅ + b sin ∅)² = a² + b²

Taking LHS 
LHS = ( a sin ∅ - b cos ∅)² + ( a cos ∅ + b sin ∅)²
= a² sin² ∅ - 2 a sin ∅ * b cos∅ + b² cos² ∅ + a² cos²∅ + 2 a cos ∅ * b sin ∅ + b² sin² ∅
= a² sin² ∅ - 2 ab sin ∅ cos ∅ + b cos²∅ + a² cos²∅ + 2 ab sin ∅ cos ∅ + b² sin² ∅
= a² sin² ∅ +b² sin²∅ + a² cos²∅ + b² cos² ∅
= a² ( sin²∅+cos²∅) + b²(sin²∅+cos²∅)
= a² ( 1 ) + b² ( 1 )
since sin² ∅ + cos² ∅ = 1
= a²+ b²
= RHS

∴ LHS = RHS
Hence, proved that
( a sin ∅ - b cos ∅)²+ ( a cos ∅ + b sin ∅)² = a² + b²

Hope this helps you !!
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