Question

Find the value of the given expression.
$3(sinx-cosx{)}^4+6(sinx+cosx{)}^2+4(si{n}^{6}x+co{s}^{6}x)$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

For ease of use, replacing x with $\theta$ :
$3\times (si{n}^2\theta +co{s}^2\theta -(2\times sin\theta \times cos\theta ){)}^2+6\times (si{n}^2\theta +co{s}^2\theta +(2\times sin\theta \times cos\theta ))+4\times (si{n}^2\theta +co{s}^2\theta )\times (si{n}^4\theta +co{s}^4\theta -(si{n}^2\theta \times co{s}^2\theta ))$
$=3\times (1-2sin\theta \times cos\theta {)}^2+6\times (1+2sin\theta \times cos\theta )+4\times (1)\times (si{n}^4\theta +co{s}^4\theta +2si{n}^2\theta \times co{s}^2\theta -3si{n}^2\theta \times co{s}^2\theta )$
$=3(1+4si{n}^2\theta \times co{s}^2\theta -4sin\theta \times cos\theta )+6+12sin\theta \times cos\theta +4\times \left(\right(si{n}^2\theta +co{s}^2\theta {)}^2-3si{n}^2\theta \times co{s}^2\theta )$
$=3+12si{n}^2\theta \times co{s}^2\theta -12\times sin\theta \times cos\theta +6+12\times sin\theta \times cos\theta +4\times (1-3si{n}^2\theta \times co{s}^2\theta )$
$=9+12si{n}^2\theta \times co{s}^2\theta +4-12si{n}^2\theta \times co{s}^2\theta$
$=13$