\(\cos ^{4} x-\sin ^{4} x=\cos (2 x)\)
\(\quad\) Hint: Will need
\(\cos (2 A)=\cos ^{2} A-\sin ^{2} A\)
\(=\left(\cos ^{2} x\right)^{2}-\left(\sin ^{2} x\right)^{2}\)
\(=\left(\cos ^{2} x+\sin ^{2} x\right)\left(\cos ^{2} x-\sin ^{2} x\right)\)
\(=(1)\left(\cos ^{2} x-\sin ^{2} x\right)\)
\(=\cos{(2x)}\)