## Simplify \(\sin ^{4} x-\cos ^{4} x+\cos ^{2} x ?\)

Answer 1:

The answer is \(=\sin ^{2} x\)

Explanation:

We need

\(\sin ^{2} x+\cos ^{2} x=1\)

\(a^{2}-b^{2}=(a+b)(a-b)\)

\(a^{4}-b^{4}=\left(a^{2}+b^{2}\right)\left(a^{2}-b^{2}\right)\)

The expression is

\(\sin ^{4} x-\cos ^{4} x+\cos ^{2} x\)

\(=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)+\cos ^{2} x\)

\(=\left(\sin ^{2} x-\cos ^{2} x\right)+\cos ^{2} x\)

\(=\sin ^{2} x\)

Answer 2:

\(\sin ^{4} x-\cos ^{4} x+\cos ^{2} x=\sin ^{2} x\)

\(=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)+\cos ^{2} x\)

\(=\sin ^{2} x-\cos ^{2} x+\cos ^{2} x\)

\(=\sin ^{2} x\)