Question

If $\;\sin \theta ,\;1,\cos 2\theta $ are in G.P., then $\theta $ is equal to $(n \in Z)$ ?
A. $n\pi + {( - 1)^n}\dfrac{\pi }{2}$
B. $n\pi + {( - 1)^{n - 1}}\dfrac{\pi }{2}$
C. $2n\pi $
D. None of these

Answer 1

Hint: In this problem, first we will compute the geometric mean of the progression given, which results in a trigonometric cubic equation. Then, we will deduce the equation into quadratic form and find the roots of the equation formulated by geometric mean previously, which will give the ultimate value of $\theta $ .
Formula used:
The formula used in this problem is: -
The geometric mean (G.M.) of G.P. $a,b,c........$ which is calculated as: ${b^2} = ac$.
Complete Step by Step Solution:
G.P. (given) = $\;\sin \theta ,\;1,\cos 2\theta \ldots \ldots .\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\; \ldots \left( 1 \right)$
We know that, for a G.P. $a,b,c........$ , the geometric mean (G.M.) is calculated as: ${b^2} = ac$
Hence, from $\left( 1 \right)$ , we get
$\sin \theta .\cos 2\theta = {(1)^2}$
Also, we have, $\cos 2\theta = 1 - 2{\sin ^2}\theta $
$ \Rightarrow \sin \theta (1 - 2{\sin ^2}\theta ) = 1$
$ \Rightarrow 2{\sin ^3}\theta - \sin \theta + 1 = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)$
To factorize the above cubic polynomial we have to first find one rational zero of this equation that is one value of \[\sin \theta \] for which the value of the equation is zero.
At \[\sin \theta =-1\]the equation will be zero. So the first root of the equation will be \[(1+\sin \theta )\].
We will now divide the whole polynomial $(2{\sin ^3}\theta - \sin \theta + 1) = 0$ by the root \[(1+\sin \theta )\] to reduce the polynomial into quadratic equation.
$\frac{2{{\sin }^{3}}\theta -\sin \theta +1}{1+\sin \theta }=2{{\sin }^{2}}\theta -2\sin \theta +1$
So,
$ \Rightarrow (1 + \sin \theta )(2{\sin ^2}\theta - 2\sin \theta + 1) = 0$
i.e., $(1 + \sin \theta ) = 0$ or $(2{\sin ^2}\theta - 2\sin \theta + 1) = 0$
But, $(2{\sin ^2}\theta - 2\sin \theta + 1)$ cannot be equal to zero as the discriminant value for this is negative. Therefore, $(2{\sin ^2}\theta - 2\sin \theta + 1) > 0$ always.
Now, Taking the case when $(1 + \sin \theta ) = 0$
$ \Rightarrow \sin \theta = - 1$
Or, we can write it as: $\sin \theta = \sin ( - \dfrac{\pi }{2})$
$ \Rightarrow \theta = n\pi + {( - 1)^n}\left( { - \dfrac{\pi }{2}} \right)$
$ \Rightarrow \theta = n\pi + {( - 1)^{n + 1}}\dfrac{\pi }{2} = n\pi + {( - 1)^{n - 1}}\dfrac{\pi }{2}$
Thus, the value of $\theta $ for the given G.P. is $n\pi + {( - 1)^{n - 1}}\dfrac{\pi }{2}$
Hence, the correct option is (B) $n\pi + {( - 1)^{n - 1}}\dfrac{\pi }{2}$
Note: Since the problem is based on geometric progression which includes trigonometric terms hence, it is essential to analyze the question very carefully, and apply the required identities to solve the question. The calculation part must be performed precisely, especially for obtaining the general solution of $\sin \theta = - 1$.