
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
164.4k+ 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
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
