Question

cos^ 6A + sin^ 6 A in terms of sin 2 A​

Answer 1

Answer:

Prove that :sin^6A+cos^6A=1 -3sin^2Acos^2A. ... sin6A+cos6A = (sin2A)3 + ( cos2A)3 = (sin2A + cos2A)3 − 3 sin2A

Answer 2

$\sf \large \mathfrak{ \color{blue}Solution : - }$


$\sf {cos}^{6} A + {sin}^{6} A = ( {cos}^{2} A {)}^{3} + ( {sin}^{2}A {)}^{3}$


$\sf \bigg[ \big( {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} \big) \bigg]$


$\sf \implies \bigg( {cos}^{2} A + {sin}^{2} A \bigg) \bigg( {cos}^{4}A - {cos}^{2} A \: {sin}^{2} A + {sin}^{4} A \bigg)$


$\sf \implies 1\bigg( {cos}^{4} A - {cos}^{2} A \: {sin}^{2} A + {sin}^{4} A \bigg)$



$\sf \implies 1\bigg( {cos}^{4} A \: {sin}^{4} A - {cos}^{2} A \: {sin}^{2} A + 2 {sin}^{2} A {cos}^{2} A - {sin}^{2}A \: {cos}^{2} A \bigg)$


$\sf \implies \bigg( \big( {cos}^{2} A + {sin}^{2} A { \big)}^{2} - 3 {sin}^{2} A \: {cos}^{2} A \bigg)$


$\sf \implies \: 1 - \large{ \frac{3}{4} } \bigg(2 \: sinA \: cosA { \bigg)}^{2}$


$\sf \color{red} \: \implies \: 1 - \large{ \frac{3}{4} } {sin}^{2} \: 2A$