Question

$2(sin^6 \theta + cos^6 \theta) -3(sin^4 + cos^4 \theta) + 1 = 0$

Prove the given, thanks!

Answer 1

To prove:

$2(sin^{6}\theta+cos^{6}\theta)-3(sin^{4}\theta+cos^{4}\theta)+1=0$

Step-by-step explanation:

$L.H.S.=2(sin^{6}\theta+cos^{6}\theta)-3(sin^{4}\theta+cos^{4}\theta)+1$

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

• since $a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)$
• and $a^{2}+b^{2}=(a+b)^{2}-2ab$

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

$=2-6\:sin^{2}\theta\:cos^{2}\theta-3+6\:sin^{2}\theta\:cos^{2}\theta+1$

$=0=R.H.S.$

Hence proved.

Note:

We must know that, $sin^{2}\theta+cos^{2}\theta=1$ when we try to solve the problem.