Question

The expression
$3[{\sin }^4(\frac{3\pi }{2}-\alpha )+{\sin }^4(3\pi +\alpha \left)\right]-2[{\sin }^6(\frac{\pi }{2}+\alpha )+{\sin }^6(5\pi -\alpha \left)\right]$ is equal to

• A. $0$
• B. $-1$
• C. $1$
• D. $3$

Answer 1

The correct option is C $1$

$3[si{n}^4(\frac{3\pi }{2}-\alpha )+si{n}^4(3\pi +\alpha \left)\right]-2[si{n}^6(\frac{\pi }{2}+\alpha )+si{n}^6(5\pi -\alpha \left)\right]$
$=3[{\cos }^4\alpha +{\sin }^4\alpha ]-2[{\cos }^6\alpha +{\sin }^6\alpha ]$
$=3\left[\right({\cos }^2\alpha {)}^2+\left({\sin }^2\alpha {)}^2\right]-2\left[\right({\cos }^2\alpha {)}^3+\left({\sin }^2\alpha {)}^3\right]$
$=3[1-2{\sin }^2\alpha {\cos }^2\alpha ]-2[1-3{\sin }^2\alpha co{s}^2\alpha ]$
$=1$