Question

Prove sin^6x+cos^6x+3 sin^2x cos^2x=1

Answer 1

its a formulae

we know (sinx+cosx)^3= sin^3x +cos^3x+3sinxcosx(sinx+cosx)

(a+b)^3= a^3 +b^3+3ab(a+b)

Explanation:

(sin^2x)^3 +cos^2x)^3+3sin^2x cos^2x(sin^2x+cos^2x)

according to formulae

(sin^2x+cos^2x)^3 {as we know sin^2x+cos^2x =1}

so the (1)^3= 1

{proved}

thank u

Answer 2

Answer: see proof below

Explanation:

Use the formula for factoring a cubic: (a³ + b³) = (a + b)(a² - ab + b²)

and the formula for a perfect square: a² + 2ab + b² = (a + b)²

and the Pythagorean Identity: cos²x + sin²x = 1

Proof LHS → RHS

Given: sin⁶x + cos⁶x + 3sin²x cos²x

Regroup: (sin²x)³ + (cos²x)³ + 3sin²x cos²x

Factor Cubic: (sin²x + cos²x)(sin⁴x - sin²x cos²x + cos⁴x) + 3sin²x cos²x

Pythagorean Identity: 1(sin⁴x - sin²x cos²x + cos⁴x) + 3sin²x cos²x

Add like terms: sin⁴x + 2sin²x cos²x + cos⁴x

Regroup: (sin²x)² + 2sin²x cos²x + (cos²x)²

Factor Perfect Square: (sin²x + cos²x)²

Pythagorean Identity: (1)²

Simplify: 1

LHS = RHS: 1 = 1
`\checkmark\\`