
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
162.3k+ views
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 $\sin \theta = - 1$.
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 $\sin \theta = - 1$.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
