Question

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

• A. $0$
• B. $1$
• C. $3$
• D. $\sin \left(4\alpha \right)+\sin \left(6\alpha \right)$

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Step-1: Simplify the given data.

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

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

Step-2: Apply, property ${\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1$.

$=3\left[1-2{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right)\right]-2\left[1\times \left({\cos }^4\left(\alpha \right)+{\sin }^4\left(\alpha \right)\right)-{\sin }^2\left(\alpha \right){\cos }^2\left(\alpha \right)\right]$

Step-3: Apply formula ${\left({\cos }^2\left(\alpha \right)+{\sin }^2\left(\alpha \right)\right)}^2-2{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right)$

$$\begin{gathered} =3\left[1-2{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right)\right]-2\left[{\left({\cos }^2\left(\alpha \right)+{\sin }^2\left(\alpha \right)\right)}^2-2{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right)-{\sin }^2\left(\alpha \right){\cos }^2\left(\alpha \right)\right] \\ =3\left[1-2{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right)\right]-2\left[1-3{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right)\right] \\ =3-6{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right)-2+6{\cos }^2\left(\alpha \right){\sin }^2\left(\alpha \right) \\ =1 \end{gathered}$$

Therefore, correct answer is option $\left(B\right)$