Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $0$

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

Consider ${\sin }^6\theta +{\cos }^6\theta$

$={\left({\sin }^2\theta \right)}^3+{\left({\cos }^2\theta \right)}^3$

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

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

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

Consider ${\sin }^4\theta +{\cos }^4\theta$

$={\left({\sin }^2\theta \right)}^2+{\left({\cos }^2\theta \right)}^2$

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

$={\left(1\right)}^2-2{\sin }^2\theta {\cos }^2\theta$ since ${\sin }^2\theta +{\cos }^2\theta =1$

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

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

$=2[1-3{\sin }^2\theta {\cos }^2\theta ]-3[1-2{\sin }^2\theta {\cos }^2\theta ]+1$

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

$=0$