${sin}^{6} A + {cos}^{6} A + 3 {sin}^{2} A {cos}^{2} A - 1 = .$

Open in App

Solution

${sin}^{6} A + {cos}^{6} A + 3 {sin}^{2} A {cos}^{2} A - 1 = ({sin}^{2} A + {cos}^{2} A) ({sin}^{4} A - {sin}^{2} A {cos}^{2} A + {cos}^{4} A + 3 {sin}^{2} A {cos}^{2} A - 1)$
$= 1 \times ({sin}^{4} A + 2 {sin}^{2} A {cos}^{2} A + {cos}^{4} A) - 1$
$= ({sin}^{2} A + c o s^{2} A)^{2} - 1$
$= (1)^{2} - 1 = 1 - 1 = 0$

Suggest Corrections

Similar questions

Prove ${sin}^{6} θ + {cos}^{6} θ = 1 - 3 {sin}^{2} θ {cos}^{2} θ$

\int \frac{tan 2 θ}{\sqrt{{cos}^{6} θ + {sin}^{6} θ}} d θ

{sin}^{6} x + {cos}^{6} x

c o s^{6} α + s i n^{6} α + K s i n^{2} 2 α = 1,

2 ({sin}^{6} A + {cos}^{6} A) - 3 ({sin}^{4} A + {cos}^{4} A) + 1 = 0.

3 [{sin}^{4} (\frac{3}{2} π - α) + {sin}^{4} (3 π + 4)] - 2 [{sin}^{6} (\frac{1}{2} π + α) + {sin}^{6} (5 π - α)]

Join BYJU'S Learning Program