Question 204685
{{{cos^2 (x) + 2sin^2 (x) - 1 =cos^2 (x) + sin^2 (x)+ sin^2 (x) - 1}}}
But, {{{sin^2(x)+cos^2(x)=1}}}. So
{{{cos^2 (x) + sin^2 (x)+ sin^2 (x) - 1=1 + sin^2(x) - 1}}}
={{{sin^2(x) + 1 -1}}}
={{{sin^2(x)}}}