Question

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

• A. 0
• B. 1
• C. 3
• D.

Answer 1

The correct option is B 1

(b) 3$[si{n}^4(\frac{3\pi }{2}-\alpha )+si{n}^4(3\pi +\alpha \left)\right]$ - $[si{n}^6(\frac{\pi }{2}+\alpha )+si{n}^6(5\pi -\alpha \left)\right]$
$=3(-cos\alpha {)}^4+(-sin\alpha {)}^4-2co{s}^6\alpha +si{n}^6\alpha$
$=3(co{s}^2\alpha +si{n}^2\alpha {)}^2-2si{n}^2\alpha co{s}^2\alpha -2(co{s}^2\alpha +si{n}^2\alpha {)}^3-3si{n}^2\alpha co{s}^2\alpha (co{s}^2\alpha +si{n}^2\alpha )$
$=3-6si{n}^2\alpha co{s}^2\alpha -2+6si{n}^2\alpha co{s}^2\alpha =3-2=1$
Trick: Put $\alpha =0$ , the value of expressions remains 1 i.e., it is independent of $\alpha$