Question

sin6theeta+cos6theeta+3sin²theetacos²theeta​

Answer 1

Answer:

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

Step-by-step explanation:

sin^6theeta= (sin^2theeta)^3

cos^6theeta= (cos^2theeta)^3

according to problem;

• multiply (sin^2theeta+cos^2theeta) to

(3sin^2theeta cos^2theeta)

• (sin^2theeta)^3+(cos^2theeta)^3+3sin^2theeta cos^2theeta( sin^2theeta+cos^2theeta)

• this is in the form of (a+b)^3

• a = sin^2theeta, b= cos^2 2theeta

• (sin^2theeta +cos^2theeta)^3

Answer 2

Answer :-

→ 1

Explanation :-

Solving $\sf{sin^6\theta+cos^6\theta}$ :-

$\sf{sin^6\theta+cos^6\theta=sin^6\theta+cos^6\theta+1-1}$

$\sf{sin^6\theta+cos^6\theta=sin^6\theta+cos^6\theta+1-(1)^3}$

$\sf{=sin^6 \theta+cos^6 \theta-(sin^2\theta+cos^2\theta)^3+1}$

$\sf{=sin^6\theta+cos^6\theta-sin^6\theta-3sin^4\theta .cos^2\theta-3sin^2\theta.cos^4\theta-sin^6\theta+1}$

$\sf{=1-3sin^2\theta .cos^2\theta (sin^2\theta +cos^2\theta )}$

$\sf{=1-3sin^2\theta .cos^2\theta }$

___________________

Now,

$\sf{sin^ 6\theta+cos^6\theta+3sin^2\theta.cos^2\theta}$

$\sf{=(\sf{1-3sin^2\theta .cos^2\theta })+3sin^2\theta .cos^2\theta}$

$\sf{=1}$

Therefore, the answer is 1.