Question

The value of $3(sinx - cosx)^4 + 6(sinx + cosx)^2 + 4(sin^6x + cos^6x) =$

• A. $11$
• B. $12$
• C. $13$
• D. $14$

Answer 1

Answer: (C)

$3\,\left(sin\, x - cos\, x\right)^{4} + 6 \,\left(sin\, x + cos\, x\right)^{2} +4\, \left(sin^{6}\, x + cos^{6}\, x\right)$
$=3\,\left(1-sin\,2x\right)^{2}+6\left(1+sin\,2x\right)+4[\left(sin^{2}\,x+cos^{2}\,x\right)^{3}$
$-3\,sin^{2}\,x\,cos^{2}\,x\left(sin^{2}\,x+cos^{2}\,x\right)]$
$= 3 \,\left(1 - 2 \,sin\, 2x + sin^{2}\, 2x\right) + 6 + 6 \,sin\, 2x$
$+4\left[1-\frac{3}{4}sin^{2}\,2x\right]$
$=13+3\,sin^{2}\,2x-3\,sin^{2}\,2x=13$.