Question

If $\sin \alpha ,{\sin }^2\alpha ,1,{\sin }^4\alpha$ and ${\sin }^5\alpha$ are in A.P. where $-\pi <a<\pi$, then $\alpha$ lies in the interval-

• A. $(\frac{-\pi }{2},\frac{\pi }{2})$
• B. $(\frac{-\pi }{3},\frac{\pi }{3})$
• C. $(\frac{-\pi }{6},\frac{\pi }{6})$
• D. none of these

Answer 1

The correct option is A $(\frac{-\pi }{2},\frac{\pi }{2})$

We have,

$\sin \alpha ,{\sin }^2\alpha ,1,{\sin }^4\alpha and{\sin }^5\alpha$ in A.P.

Then,

$\text{First}\text{term}a=\sin \alpha$

$\text{Common}\text{difference}\text{=}\text{Second}\text{term}\text{-}\text{first}\text{term}$

Where

$T_1=Firstterm$

$T_2=Secondterm$

$T_3=Thirdterm$

$.......$

Then, we know that,

If the series in an A.P.

$T_2-T_1=T_3-T_2=T_4-T_3=T_5-T_4$

${\sin }^2\alpha -\sin \alpha =1-{\sin }^2\alpha ={\sin }^4\alpha -1={\sin }^5\alpha -{\sin }^4\alpha$

$\Rightarrow {\sin }^2\alpha -\sin \alpha =1-{\sin }^2\alpha$

$\Rightarrow {\sin }^2\alpha +{\sin }^2\alpha -\sin \alpha =1$

$\Rightarrow 2{\sin }^2\alpha -\sin \alpha -1=0$

$\Rightarrow 2{\sin }^2\alpha -(2-1)\sin \alpha -1=0$

$\Rightarrow 2{\sin }^2\alpha -2\sin \alpha +\sin \alpha -1=0$

$\Rightarrow 2\sin \alpha (\sin \alpha -1)+1(\sin \alpha -1)=0$

$\Rightarrow (\sin \alpha -1)(2\sin \alpha +1)=0$

$\Rightarrow \sin \alpha -1=0,2\sin \alpha +1=0$

$\Rightarrow \sin \alpha =1,\sin \alpha =\frac{-1}{2}$

$\Rightarrow \sin \alpha =\sin \frac{\pi }{2},\sin \alpha =-\sin \frac{\pi }{6}$

$\Rightarrow \alpha =\frac{\pi }{2},\sin \alpha =\sin (\pi +\frac{\pi }{6})\because \sin (\pi +\theta )=-\sin \theta$

$\Rightarrow \alpha =\frac{\pi }{2},\alpha =\frac{7\pi }{6}$

Similarly we can show that,

$\alpha =-\frac{\pi }{2}$

Hence, $\alpha =(-\frac{\pi }{2},\frac{\pi }{2})$

Hence, this is the required answer.