Question

cos^6A-sin^6A=cos2A(1-1/4sin^2A),PROVE IT

Answer 1

$cos^6A-sin^6A$

$=(cos^2A)^3-(sin^2A)^3$

$\text{Using}$

$\boxed{\bf\,a^3-b^3=(a-b)(a^2+ab+b^2)}$

$=(cos^2A-sin^2A)(cos^4A+cos^2A\,sin^2A+sin^4A)$

$\text{Using}$

$\boxed{\bf\,cos2A=cos^2A-sin^2A}$

$=cos2A[(cos^2A)^2+(sin^2A)^2+2sin^2A\,cos^2A-sin^2A\,cos^2A]$

$=cos2A[(cos^2A+sin^2A)^2-(sin\,A\,cos\,A)^2]$

$\text{Using}$

$\boxed{\bf\,cos^2A+sin^2A=1}$

$=cos2A[1-{\frac{1}{4}}\,(2sinA\,cosA)^2]$

$\text{Using}$

$\boxed{\bf\,sin\,2A=2\,sin\,A\,cos\,A}$

$=cos2A[1-{\frac{1}{4}}\,sin^2\,2A]$

$\implies\boxed{\bf\,cos^6A-sin^6A=cos2A[1-{\frac{1}{4}}\,sin^2\,2A]}$

Answer 2

Step-by-step explanation:

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