Question

Prove that cos 4x = 1 - 8 sin${}^{2}x{\text{cos}}^{2}x$.

Answer 1

We have

LHS = $cos4x$

= $cos2\left(2x\right)=cos2\theta$, where $2x=\theta$

= ${\text{cos}}^2\theta -{\text{sin}}^2\theta$

$[\because \text{cos}2\theta ={\text{cos}}^2\theta -{\text{sin}}^2\theta ]$

= $(1-2{\text{sin}}^2\theta )=\{1-2(\text{sin}2x{)}^2\}[\because \text{sin}2x=2\text{sin}x\text{cos}x]$

=$1-8{\text{sin}}^{2}x{\text{cos}}^{2}x=\text{RHS}.$

$\text{Hence, cos}4x=1-8{\text{sin}}^{2}x{\text{cos}}^{2}x$