Question

Trigo Trigo :p

Prove the identity :

$( { \sin }^{8} \beta - { \cos }^{8} \beta ) = ( { \sin}^{2} \beta - { \cos}^{2} \beta )(1 - 2 { \sin}^{2} \beta { \cos}^{2} \beta )$




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Answer 1

Answer:

$LHS = sin^{8}\beta - cos ^{8}\beta \\=(sin^{4}\beta^{2}-(cos^{4}\beta)^{2}\\=(sin^{4}\beta+cos^{4}\beta)(sin^{4}\beta-cos^{4}\beta)</p><p>$

/* By algebraic identity:

a²-b² = (a+b)(a-b) */

$= [(sin^{2}\beta+cos^{2}\beta)^{2}-2sin^{2}\beta cos^{2}\beta][(sin^{2}\beta)^{2}-(cos^{2}\beta)^{2}]$

/* we know that,

a²+b² = (a+b)² - 2ab Or

a⁴+b⁴ = (a²+b²)²-2a²b² */

/* sin²A + cos²A = 1 */

$= (1-2sin^{2}\beta cos^{2}\beta)(sin^{2}\beta-cos^{2}\beta)(sin^{2}\beta+cos^{2}\beta)$

$=(1-2sin^{2}\beta cos^{2}\beta)(sin^{2}\beta-cos^{2}\beta)\\=RHS$

•••♪

Answer 2

Answer:

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