Question

Prove that:

$co{s}^{6}A-si{n}^{6}A=cos2A(1-\frac{1}{4}si{n}^{2}2A)$

Answer 1

$\text{LHS}=co{s}^{6}A-si{n}^{6}A=(co{s}^{2}A{)}^3-(si{n}^{2}A{)}^3=(co{s}^{2}A-si{n}^{2}A)(co{s}^{4}A+si{n}^{2}A.co{s}^{2}A+si{n}^{4}A)[\because a^3-b^3=(a-b\left)\right(a^2+ab+b^2\left)\right]=cos2A(co{s}^{4}A+2si{n}^{2}Aco{s}^{2}A+si{n}^{4}A-si{n}^{2}Aco{s}^{2}A)[\because co{s}^{2}A-si{n}^{2}A=co{s}^{2}A\text{and Adding and sutrading}si{n}^{2}co{s}^{2}A]=cos2A\left[\right(si{n}^{2}A+co{s}^{2}A{)}^2-\frac{4}{4}si{n}^{2}Aco{s}^{2}A]$

$=cos2A[1-\frac{1}{4}(2sinAcosA{)}^2]=cos2A[1-\frac{1}{4}si{n}^{2}2A]$

= RHS