2011/078) If x^2+y^2=1,then (4x^3-3x)^2+(3y-4y^3)^2=

This can be solved both algebraically and also using trigonometry.

Algebra method

We putting y ^2 = (-1x^2) that is converting to terms of x

x² * (4x² - 3)² + y² * (3 - 4y²)²

=x² * (4x² - 3)² + (1 - x²) * [3 - (4 - 4x²)]²
= x² * (4x² - 3)² - (x² - 1) * ( 4x² - 1)²
= x² * (4x² - 3)² - x² * ( 4x² - 1)² + ( 4x² - 1)²
= x² [ (4x² - 3)² - ( 4x² - 1)² ]+ ( 4x² - 1)²
= x² [ (4x² - 3 + 4x² - 1)(4x² - 3 - 4x² + 1) ] + ( 4x² - 1)²
= x² * (8x² - 4)(-2) + ( 4x² - 1)²
= -16x^4 + 8x² + 16x^4 - 8x² + 1

= 1

Now for trigonometric method

Given : x² + y² = 1
then x can be considered as sin θ and y as cos θ
{Since sin² θ + cos² θ = 1 }

Now, 4x³ - 3x = 4 sin³ θ - 3 sin θ = sin 3θ
and 3y - 4y³ = 3 cos θ - 4 cos³ θ = cos 3θ

=> (4x³ - 3x)² + (3y - 4y³)²
= sin² 3θ + cos² 3θ
= 1