Prove the trigonometric identity ?

sin^2alpha (4cos^2alpha-1)^2+cos^2alpha(4sin^2alpha-1)^2=1

1 Answer
Jun 3, 2018

LHS=sin^2x(4cos^2x-1)^2+cos^2x(4sin^2x-1)^2

=sin^2x[16cos^4x-8cos^2x+1]+cos^2x[16sin^4x-8sin^2x+1]

=16cos^4x*sin^2x-8cos^2x*sin^2x+sin^2x+16sin^4x*cos^2x-8sin^2xcos^2x+cos^2x

=sin^2x+cos^2x+16cos^4x*sin^2x+16sin^4x*cos^2x-16sin^2x*cos^2x

=1+16sin^2xcos^2x[cos^2x+sin^2x-1]

=1+16sin^2xcos^2x[1-1]

=1+16sin^2xcos^2(x)xx0=1=RHS