Question

Prove that sin^6 theta + cos^6 theta+ 3 sin^2 theta cos^2 theta = 1.

Answer 1

Answer:

Hello mate!

Here's the solution.

sin²θ + cos²θ = 1

Therefore (sin²θ + cos²θ)³ = 1

or

(sin²θ)³ + (cos²θ)³ + 3sin²θ cos²θ(sin²θ + cos²θ) = 1

or,

sin^6θ + cos^6θ+ 3sin²θ cos²θ = 1

HOPE IT HELPS YOU ✌

Answer 2

$\huge \sf {\orange {\underline {\pink{\underline{Answer :-}}}}}$

To prove that sin^6 theta + cos^6 theta+ 3 sin^2 theta cos^2 theta = 1.

$\sf\underline\red{Solution:}$

$\rm\leadsto{sin²θ + cos²θ = 1}$

$\rm\leadsto{Therefore (sin²θ + cos²θ)³ = 1}$

$\rm{or ,}$

$\rm\leadsto{(sin²θ)³ + (cos²θ)³ +}$

$\rm\leadsto{ 3sin²θ cos²θ(sin²θ + cos²θ) = 1}$

$\rm{or,}$

$\rm\leadsto{sin^6θ + cos^6θ+ }$

$\rm\leadsto{3sin²θ cos²θ = 1}$

${\huge{\underline{\small{\mathbb{\pink{HOPE\:HELPS\:UH :)}}}}}}$

$\red{\tt{sωєєтcнαям♡~}}$