Question

$3\left[{\sin }^4(\frac{3\pi }{2}-\alpha )\right]+\left[{\sin }^4\right(3\pi +\alpha \left)\right]-2[{\sin }^6(\frac{\pi }{2}+\alpha )+{\sin }^6(5\pi -\alpha \left)\right]=$

• A. $0$
• B. $1$
• C. $3$
• D. $\sin 4\alpha +\cos 6\alpha$

Answer 1

The correct option is B $1$

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

Therefore, Answer is $1$