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