Question

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

Answer 1

Answer:

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)

Step-by-step 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

Step-by-step 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    $\checkmark$