Question

PROVE THE FOLLOWING IDENTITY.........

${sin}^{6} \alpha + {cos}^{6} \alpha = 1 - {sin}^{2} \ \alpha \: {cos}^{2} \alpha$
NEEDED URGENTLY
THE CORRECT ANSWER WILL BE MARKED BRAINLEST​

Answer 1

(instead of α , I use A)

Answer:

Step-by-step explanation:

Given :

To prove :

$sin^6A + cos^6A = 1-3sin^2Acos^2A$ (here, you missed 3)

Solution :

We know that,

⇒ sin² A = cos² A = 1,

⇒ a³ + b³ = (a + b)(a² - ab + b²)

⇒ (a ± b)² = a² ± 2ab + b²

LHS = $sin^6A + cos^6A$

⇒ $(sin^2A)^3 + (cos^2A)^3$

⇒ $(sin^2A + cos^2A)([sin^2A)^2 - sin^2Acos^2A + [cos^2A]^2)$

⇒ $(1)(sin^4A - sin^2Acos^2A + cos^4A)$

⇒ $(sin^4A + cos^4A) - sin^2Acos^2A$

⇒ $([sin^2A]^2 + [cos^2A]^2) - sin^2Acos^2A$

⇒ $([sin^2A]^2 + [cos^2A]^2 + 2sin^2Acos^2A) - 2sin^2Acos^2A - sin^2Acos^2A$

⇒ $(sin^2A+ cos^2A)^2 - 2sin^2Acos^2A - sin^2Acos^2A$

⇒ $(1)^2 - 3sin^2Acos^2A$

⇒ $1 - 3sin^2Acos^2A$ = RHS

Hence, proved,.

Proof :

let A = 60°

⇒ LHS = $sin^6A + cos^6A = (\frac{\sqrt{3}}{2})^6 + (\frac{1}{2})^6 = \frac{27}{64} + \frac{1}{64} = \frac{28}{64}=\frac{7}{16}$

⇒ RHS = $1 - 3sin^2Acos^2A = 1 - 3(\frac{\sqrt{3}}{2})^2(\frac{1}{2})^2 = 1- 3(\frac{3}{4})(\frac{1}{4}) = 1 - \frac{9}{16} = \frac{16-9}{16} = \frac{7}{16}$

LHS = RHS