
The roots of the equation\[\]\[1 - cos\theta = sin\theta \]. \[sin\dfrac{\theta }{2}\]is
A. \[k\pi ,k \in I\]
B. \[2k\pi ,k \in I\]
C. \[k\pi 2,k \in I\]
D. none of these
Answer
220.8k+ views
Hint: To solve this question, we will use the formula from trigonometry \[\cos \theta \]and \[\sin 2\theta \]. we will derive the value of \[k\]and the \[\theta \].also we will simplify the equation using trigonometry formulas and get the resultant values. The relationship between the length of the sides of the right triangle and the measurement of the angles is known as the trigonometric ratio. Formulas involving trigonometric functions are known as trigonometric idents. For each possible value of the variables, these identities hold true.
Formula Used: The trigonometric formula of cos and sin are
\[1 - \cos \dfrac{\theta }{2}\]
\[\sin \theta = 2\sin \dfrac{\theta }{2}.\cos \dfrac{\theta }{2}\]
Complete step by step solution: The roots of the equation\[\]\[1 - cos\theta = sin\theta \]. \[sin\dfrac{\theta }{2}\]
By using the trigonometry formula
\[1 - cos\theta = sin\theta \]. \[sin\dfrac{\theta }{2}\]
\[2{\sin ^2}\dfrac{\theta }{2} = \]\[2\sin \dfrac{\theta }{2}.\cos \dfrac{\theta }{2}.\sin \dfrac{\theta }{2}\]
Simplify 2 on both sides
\[{\sin ^2}\theta = {\sin ^2}\dfrac{\theta }{2}.\cos \dfrac{\theta }{2}\]
\[{\sin ^2}\theta - {\sin ^2}\dfrac{\theta }{2}.\cos \dfrac{\theta }{2} = 0\]
Take \[\sin \theta \] common
\[{\sin ^2}\dfrac{\theta }{2}(1 - \cos \dfrac{\theta }{2}) = 0\]
Using trigonometry formula
\[1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}\]
Divide by \[\dfrac{\theta }{2}\]
\[1 - \cos \dfrac{\theta }{2} = 2{\sin ^2}\dfrac{\theta }{4}\]
Using this formula in the equation
\[{\sin ^2}\dfrac{\theta }{2}(1 - \cos \dfrac{\theta }{2}) = 0\]
\[{\sin ^2}\dfrac{\theta }{2}.2{\sin ^2}\dfrac{\theta }{4} = 0\]
Take \[2\] to the other side
\[{\sin ^2}\dfrac{\theta }{2}.{\sin ^2}\dfrac{\theta }{4} = 0\]
\[{\sin ^2}\dfrac{\theta }{2} = 0\] or \[{\sin ^2}\dfrac{\theta }{4} = 0\]
Taking square root on both sides
\[\sin \dfrac{\theta }{2} = 0\] or \[\sin \dfrac{\theta }{4} = 0\]
In trigonometry, if \[\sin \theta = 0\] then \[\theta = k\pi \]where \[k \in i\]
Then \[\dfrac{\theta }{2} = k\pi \] or \[\dfrac{\theta }{4} = k\pi \] where \[k \in i\]
\[\theta = 2k\pi \] or \[\theta = 4k\pi \]where \[k \in i\]
Option ‘B’ is correct
Note: The trigonometry formula should be used correctly, and we have to modify the formula according to our need then the formula should vary. If we take square root on both sides, then it should be noted carefully. Three fundamental trigonometric operations are Sin, Cos, and Tan. The problem utilises trigonometric functions like sine, cosine, tangent, cotangent, cosecant, and secant. Trigonometric formulas are utilised to evaluate the problem. We can resolve issues with the angles and sides of a right triangle by using the various trigonometric identities.
Formula Used: The trigonometric formula of cos and sin are
\[1 - \cos \dfrac{\theta }{2}\]
\[\sin \theta = 2\sin \dfrac{\theta }{2}.\cos \dfrac{\theta }{2}\]
Complete step by step solution: The roots of the equation\[\]\[1 - cos\theta = sin\theta \]. \[sin\dfrac{\theta }{2}\]
By using the trigonometry formula
\[1 - cos\theta = sin\theta \]. \[sin\dfrac{\theta }{2}\]
\[2{\sin ^2}\dfrac{\theta }{2} = \]\[2\sin \dfrac{\theta }{2}.\cos \dfrac{\theta }{2}.\sin \dfrac{\theta }{2}\]
Simplify 2 on both sides
\[{\sin ^2}\theta = {\sin ^2}\dfrac{\theta }{2}.\cos \dfrac{\theta }{2}\]
\[{\sin ^2}\theta - {\sin ^2}\dfrac{\theta }{2}.\cos \dfrac{\theta }{2} = 0\]
Take \[\sin \theta \] common
\[{\sin ^2}\dfrac{\theta }{2}(1 - \cos \dfrac{\theta }{2}) = 0\]
Using trigonometry formula
\[1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}\]
Divide by \[\dfrac{\theta }{2}\]
\[1 - \cos \dfrac{\theta }{2} = 2{\sin ^2}\dfrac{\theta }{4}\]
Using this formula in the equation
\[{\sin ^2}\dfrac{\theta }{2}(1 - \cos \dfrac{\theta }{2}) = 0\]
\[{\sin ^2}\dfrac{\theta }{2}.2{\sin ^2}\dfrac{\theta }{4} = 0\]
Take \[2\] to the other side
\[{\sin ^2}\dfrac{\theta }{2}.{\sin ^2}\dfrac{\theta }{4} = 0\]
\[{\sin ^2}\dfrac{\theta }{2} = 0\] or \[{\sin ^2}\dfrac{\theta }{4} = 0\]
Taking square root on both sides
\[\sin \dfrac{\theta }{2} = 0\] or \[\sin \dfrac{\theta }{4} = 0\]
In trigonometry, if \[\sin \theta = 0\] then \[\theta = k\pi \]where \[k \in i\]
Then \[\dfrac{\theta }{2} = k\pi \] or \[\dfrac{\theta }{4} = k\pi \] where \[k \in i\]
\[\theta = 2k\pi \] or \[\theta = 4k\pi \]where \[k \in i\]
Option ‘B’ is correct
Note: The trigonometry formula should be used correctly, and we have to modify the formula according to our need then the formula should vary. If we take square root on both sides, then it should be noted carefully. Three fundamental trigonometric operations are Sin, Cos, and Tan. The problem utilises trigonometric functions like sine, cosine, tangent, cotangent, cosecant, and secant. Trigonometric formulas are utilised to evaluate the problem. We can resolve issues with the angles and sides of a right triangle by using the various trigonometric identities.
Recently Updated Pages
The maximum number of equivalence relations on the-class-11-maths-JEE_Main

A train is going from London to Cambridge stops at class 11 maths JEE_Main

Find the reminder when 798 is divided by 5 class 11 maths JEE_Main

An aeroplane left 50 minutes later than its schedu-class-11-maths-JEE_Main

A man on the top of a vertical observation tower o-class-11-maths-JEE_Main

In an election there are 8 candidates out of which class 11 maths JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Degree of Dissociation: Meaning, Formula, Calculation & Uses

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

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

