Question 1190028
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I'll be using the pythagorean trig identity {{{sin^2(x)+cos^2(x) = 1}}}
That rearranges into {{{sin^2(x) = 1-cos^2(x)}}}


From there, we can have this set of steps


{{{sin^4(x) - sin^2(x)}}}


{{{(sin^2(x))^2 - sin^2(x)}}}


{{{(1-cos^2(x))^2 - (1-cos^2(x))}}} Each sin^2 term is replaced with 1-cos^2


{{{1-2cos^2(x)+(cos^2(x))^2 - (1-cos^2(x))}}} FOIL rule


{{{1-2cos^2(x)+cos^4(x) - 1+cos^2(x)}}}


{{{cos^4(x)-cos^2(x)}}}


Therefore, {{{sin^4(x) - sin^2(x) = cos^4(x)-cos^2(x)}}} is an identity. It's a true equation for all real numbers. 
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