If the real part of the complex number $(1 - \cos \theta + 2 i\sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in (0, \pi)$, then the value of the integral $\int_{0}^{\theta} \sin x dx$ is equal to: Options: (A) 1 (B) 2 (C) -1 (D) 0

### Analyzing the Setup We are given the complex number $z = (1 - \cos \theta + 2i \sin \theta)^{-1}$ and informed that its real part is exactly $\frac{1}{5}$. Our objective is to evaluate the definite integral $\int_{0}^{\theta} \sin x \, dx$. ### The Art of Simplification The denominator $1 - \cos \theta + 2i \sin \theta$ can be simplified using standard trigonometric identities. Recall that $1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}$ and $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$. Substituting these into the expression, we obtain: $$z = \frac{1}{2 \sin^2 \frac{\theta}{2} + 4i \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$$ Factoring out $2 \sin \frac{\theta}{2}$ from the denominator yields: $$z = \frac{1}{2 \sin \frac{\theta}{2} \left( \sin \frac{\theta}{2} + 2i \cos \frac{\theta}{2} \right)}$$ ### The Rescue Mission To isolate the real part, we rationalize the denominator by multiplying the numerator and denominator by the complex conjugate, $\sin \frac{\theta}{2} - 2i \cos \frac{\theta}{2}$. The denominator becomes: $$2 \sin \frac{\theta}{2} \left( \sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2} \right)$$ The expression for $z$ is now: $$z = \frac{\sin \frac{\theta}{2} - 2i \cos \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \left( \sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2} \right)}$$ ### The Trigonometric Bridge Extracting the real part of $z$, we cancel the $\sin \frac{\theta}{2}$ term: $$\text{Re}(z) = \frac{1}{2 \left( \sin^2 \frac{\theta}{2} + 4 \cos^2 \frac{\theta}{2} \right)} = \frac{1}{5}$$ This implies $2 \sin^2 \frac{\theta}{2} + 8 \cos^2 \frac{\theta}{2} = 5$. Using the identity $\sin^2 \frac{\theta}{2} = 1 - \cos^2 \frac{\theta}{2}$, we substitute: $$2(1 - \cos^2 \frac{\theta}{2}) + 8 \cos^2 \frac{\theta}{2} = 5$$ $$6 \cos^2 \frac{\theta}{2} = 3 \Rightarrow \cos^2 \frac{\theta}{2} = \frac{1}{2}$$ For $\theta \in (0, \pi)$, this results in $\frac{\theta}{2} = \frac{\pi}{4}$, which means **$\theta = \frac{\pi}{2}$**. ### The Calculus Finale We now evaluate the integral with the determined value of $\theta$: $$\int_{0}^{\pi/2} \sin x \, dx = \left[ -\cos x \right]_{0}^{\pi/2}$$ Calculating the limits: $$-(\cos \frac{\pi}{2} - \cos 0) = -(0 - 1) = 1$$ The final value of the integral is **$1$**.

Why is $\cos \frac{\theta}{2}$ positive in our final step?

Great observation! We are given the constraint $\theta \in (0, \pi)$. This implies that $\frac{\theta}{2} \in (0, \frac{\pi}{2})$. Since $\frac{\theta}{2}$ lies in the first quadrant, all trigonometric functions, including cosine, must be positive. That is why we reject the negative root.

JEE Main 2021 (20 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: If the real part of the complex number $(1 - cos θ + 2 i sin θ)^{- 1}$ is $\frac{1}{5}$ for $θ \in (0, π)$ , then the value of the integral $\int_{0}^{θ} sin x d x$ is equal to:

Q: How do I know which half-angle identity to use? There are so many!

Look at the structure of the expression. The term $1 - \cos \theta$ is a massive hint. Whenever you see $1 \pm \cos \theta$, your first instinct should be to use the half-angle identities ($2 \sin^2 \frac{\theta}{2}$ or $2 \cos^2 \frac{\theta}{2}$). They are designed to collapse these terms into a single, manageable square.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Do not be intimidated by the inverse power. Focus on the expression inside the parentheses:

1 - cos θ + 2 i sin θ

. This is a classic setup for half-angle identities. Before you try to invert it, simplify the internal structure using

1 - cos θ = 2 sin^{2} \frac{θ}{2}

and

sin θ = 2 sin \frac{θ}{2} cos \frac{θ}{2}

Hint 2

Hint 2: The Arsenal. Once you have simplified the denominator, you will have a complex number in the form

1/ w

. To isolate the real part, you must rationalize the denominator. Multiply both the numerator and the denominator by the complex conjugate of the denominator. Remember, for any complex number

w = a + bi

, the product

w \overset{w}{ˉ} = a^{2} + b^{2}

Hint 3

Hint 3: The Execution. After rationalizing, you will find that the real part of

z

\frac{1}{2 ( s i n ^{2} \frac{θ}{2} + 4 c o s ^{2} \frac{θ}{2} )}

. Equate this to

\frac{1}{5}

and solve the resulting trigonometric equation for

cos^{2} \frac{θ}{2}

. Use the domain

θ \in (0, π)

to determine the unique value of

θ

Visualized Solution

Problem Overview

Given: $z = (1 - cos θ + 2 i sin θ)^{- 1}$
Condition: $Re (z) = \frac{1}{5}$ for $θ \in (0, π)$
Goal: Evaluate $\int_{0}^{θ} sin x d x$

Trigonometric Half-Angle Identities

To simplify $z$ , we use half-angle formulas.
$1 - cos θ = 2 sin^{2} \frac{θ}{2}$
$sin θ = 2 sin \frac{θ}{2} cos \frac{θ}{2}$

Substituting Identities into $z$

Substitute the identities into the expression for $z$ :
$z = \frac{1}{2 s i n ^{2} \frac{θ}{2} + 2 i ( 2 s i n \frac{θ}{2} c o s \frac{θ}{2} )}$
$z = \frac{1}{2 s i n ^{2} \frac{θ}{2} + 4 i s i n \frac{θ}{2} c o s \frac{θ}{2}}$

Factoring the Denominator

Notice the common term $2 sin \frac{θ}{2}$ in the denominator.
Factor it out:
$z = \frac{1}{2 s i n \frac{θ}{2} ( s i n \frac{θ}{2} + 2 i c o s \frac{θ}{2} )}$

Rationalizing the Complex Expression

To find the real part, we must remove $i$ from the denominator.
Multiply numerator and denominator by the conjugate:
Conjugate of $(sin \frac{θ}{2} + 2 i cos \frac{θ}{2})$ is $(sin \frac{θ}{2} - 2 i cos \frac{θ}{2})$

Simplifying the Denominator

Use the identity $(a + bi) (a - bi) = a^{2} + b^{2}$ :
$(sin \frac{θ}{2} + 2 i cos \frac{θ}{2}) (sin \frac{θ}{2} - 2 i cos \frac{θ}{2}) = sin^{2} \frac{θ}{2} + 4 cos^{2} \frac{θ}{2}$
The full denominator becomes: $2 sin \frac{θ}{2} (sin^{2} \frac{θ}{2} + 4 cos^{2} \frac{θ}{2})$

Extracting the Real Part

The numerator is $sin \frac{θ}{2} - 2 i cos \frac{θ}{2}$ .
Separate the real and imaginary parts.
$Re (z) = \frac{s i n \frac{θ}{2}}{2 s i n \frac{θ}{2} ( s i n ^{2} \frac{θ}{2} + 4 c o s ^{2} \frac{θ}{2} )}$
Cancel $sin \frac{θ}{2}$ :
$Re (z) = \frac{1}{2 ( s i n ^{2} \frac{θ}{2} + 4 c o s ^{2} \frac{θ}{2} )}$

Equating Real Part to $1/5$

We are given $Re (z) = \frac{1}{5}$ .
$\frac{1}{2 ( s i n ^{2} \frac{θ}{2} + 4 c o s ^{2} \frac{θ}{2} )} = \frac{1}{5}$
Cross-multiply to get:
$2 (sin^{2} \frac{θ}{2} + 4 cos^{2} \frac{θ}{2}) = 5$

Solving the Trigonometric Equation

Convert everything to cosine using $sin^{2} \frac{θ}{2} = 1 - cos^{2} \frac{θ}{2}$ :
$2 (1 - cos^{2} \frac{θ}{2} + 4 cos^{2} \frac{θ}{2}) = 5$
$2 (1 + 3 cos^{2} \frac{θ}{2}) = 5$
$2 + 6 cos^{2} \frac{θ}{2} = 5 ⟹ 6 cos^{2} \frac{θ}{2} = 3$
$cos^{2} \frac{θ}{2} = \frac{1}{2}$

Finding the Value of $θ$

We have $cos^{2} \frac{θ}{2} = \frac{1}{2}$ , so $cos \frac{θ}{2} = \pm \frac{1}{2}$ .
Given constraint: $θ \in (0, π) ⟹ \frac{θ}{2} \in (0, \frac{π}{2})$ .
In the first quadrant, cosine is positive: $cos \frac{θ}{2} = \frac{1}{2}$ .
Therefore, $\frac{θ}{2} = \frac{π}{4} ⟹ θ = \frac{π}{2}$ .

Setting Up the Definite Integral

Now substitute $θ = \frac{π}{2}$ into the original integral.
The integral becomes: $\int_{0}^{π /2} sin x d x$
Geometrically, this represents the area under the sine curve from $x = 0$ to $x = \frac{π}{2}$ .

Evaluating the Integral

Antiderivative of $sin x$ is $- cos x$ .
Apply limits from $0$ to $\frac{π}{2}$ :
$[- cos x]_{0}^{π /2} = - (cos \frac{π}{2} - cos 0)$
Substitute values: $cos \frac{π}{2} = 0$ and $cos 0 = 1$ .
$- (0 - 1) = 1$

Final Conclusion

The value of the integral $\int_{0}^{θ} sin x d x$ is $1$ .
Key Steps Reviewed:
1. Simplified $z$ using half-angle formulas.
2. Rationalized to extract $Re (z)$ .
3. Solved for $θ = \frac{π}{2}$ .
4. Evaluated the area under the sine curve.

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Analyzing the Setup

We are given the complex number

z = (1 - cos θ + 2 i sin θ)^{- 1}

and informed that its real part is exactly

\frac{1}{5}

. Our objective is to evaluate the definite integral

\int_{0}^{θ} sin x d x

The Art of Simplification

The denominator

1 - cos θ + 2 i sin θ

can be simplified using standard trigonometric identities. Recall that

1 - cos θ = 2 sin^{2} \frac{θ}{2}

and

sin θ = 2 sin \frac{θ}{2} cos \frac{θ}{2}

Substituting these into the expression, we obtain:

z = \frac{1}{2 sin ^{2} \frac{θ}{2} + 4 i sin \frac{θ}{2} cos \frac{θ}{2}}

Factoring out

2 sin \frac{θ}{2}

from the denominator yields:

z = \frac{1}{2 sin \frac{θ}{2} ( sin \frac{θ}{2} + 2 i cos \frac{θ}{2} )}

The Rescue Mission

To isolate the real part, we rationalize the denominator by multiplying the numerator and denominator by the complex conjugate,

sin \frac{θ}{2} - 2 i cos \frac{θ}{2}

The denominator becomes:

2 sin \frac{θ}{2} (sin^{2} \frac{θ}{2} + 4 cos^{2} \frac{θ}{2})

The expression for

z

is now:

z = \frac{sin \frac{θ}{2} - 2 i cos \frac{θ}{2}}{2 sin \frac{θ}{2} ( sin ^{2} \frac{θ}{2} + 4 cos ^{2} \frac{θ}{2} )}

The Trigonometric Bridge

Extracting the real part of

z

, we cancel the

sin \frac{θ}{2}

term:

Re (z) = \frac{1}{2 ( sin ^{2} \frac{θ}{2} + 4 cos ^{2} \frac{θ}{2} )} = \frac{1}{5}

This implies

2 sin^{2} \frac{θ}{2} + 8 cos^{2} \frac{θ}{2} = 5

. Using the identity

sin^{2} \frac{θ}{2} = 1 - cos^{2} \frac{θ}{2}

, we substitute:

2 (1 - cos^{2} \frac{θ}{2}) + 8 cos^{2} \frac{θ}{2} = 5

6 cos^{2} \frac{θ}{2} = 3 \Rightarrow cos^{2} \frac{θ}{2} = \frac{1}{2}

For

θ \in (0, π)

, this results in

\frac{θ}{2} = \frac{π}{4}

, which means $θ = \frac{π}{2}$ .

The Calculus Finale

We now evaluate the integral with the determined value of

θ

\int_{0}^{π /2} sin x d x = [- cos x]_{0}^{π /2}

Calculating the limits:

- (cos \frac{π}{2} - cos 0) = - (0 - 1) = 1

The final value of the integral is $1$ .

Common Traps & FAQs

Wait, why can't we just take the reciprocal of the real part of the denominator to find the real part of the whole expression?

That is a very common trap! Remember that for a complex number

z = a + bi

, the real part of

1/ z

is not simply

1/ a

. You must rationalize the denominator by multiplying by the conjugate to separate the real and imaginary components. Only then can you identify the real part correctly.

How do I know which half-angle identity to use? There are so many!

Look at the structure of the expression. The term

1 - cos θ

is a massive hint. Whenever you see

1 \pm cos θ

, your first instinct should be to use the half-angle identities (

2 sin^{2} \frac{θ}{2}

2 cos^{2} \frac{θ}{2}

). They are designed to collapse these terms into a single, manageable square.

Why is

cos \frac{θ}{2}

positive in our final step?

Great observation! We are given the constraint

θ \in (0, π)

. This implies that

\frac{θ}{2} \in (0, \frac{π}{2})

. Since

\frac{θ}{2}

lies in the first quadrant, all trigonometric functions, including cosine, must be positive. That is why we reject the negative root.

Animated Solution for Mathematics - Complex Numbers: If the real part of the complex number $(1 - cos θ + 2 i sin θ)^{- 1}$ is $\frac{1}{5}$ for $θ \in (0, π)$ , then the value of the integral $\int_{0}^{θ} sin x d x$ is equal to:

Select Answer:

Visualized Solution

Problem Overview

Trigonometric Half-Angle Identities

Substituting Identities into $z$

Factoring the Denominator

Rationalizing the Complex Expression

Simplifying the Denominator

Extracting the Real Part

Equating Real Part to $1/5$

Solving the Trigonometric Equation

Finding the Value of $θ$

Setting Up the Definite Integral

Evaluating the Integral

Final Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Art of Simplification

The Rescue Mission

The Trigonometric Bridge

The Calculus Finale

Similar Questions

The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i c o s θ}{1 - 2 i c o s θ}$ is purely imaginary, is equal

Let $A = {θ \in (- π /2, π) : \frac{3 + 2 i s i n θ}{1 - 2 i s i n θ} is purely imaginary}$ . Then the sum of the elements in A is :

If the expression $[\frac{sin ( \frac{x}{2} ) + cos ( \frac{x}{2} ) + i tan ( x )}{1 + 2 i sin ( \frac{x}{2} )}]$ is real, then the set of all possible values of $x$ is .........

Let $A = {θ \in (0, 2 π) : \frac{1 + 2 i s i n θ}{1 - i s i n θ} is purely imaginary}$ . Then the sum of the elements in $A$ is

Express $\frac{1}{1 - cos θ + 2 i sin θ}$ in the form $x + i y$ .

If $\frac{3 + i s i n θ}{4 - i c o s θ}, θ \in [0, 2 π]$ , is a real number, then the argument of $sin θ + i cos θ$ is :

If $\frac{3 + i s i n θ}{4 - i c o s θ}$ , $θ \in [0, 2 π]$ , is a real number, then an argument of $sin θ + i cos θ$ is:

A value of $θ$ for which $\frac{2 + 3 i s i n θ}{1 - 2 i s i n θ}$ is purely imaginary, is

The value of the sum $\sum_{n = 1}^{13} (i^{n} + i^{n + 1})$ , where $i = - 1$ , equals

Select Answer:

Visualized Solution

Problem Overview

Trigonometric Half-Angle Identities

Substituting Identities into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Factoring the Denominator

Rationalizing the Complex Expression

Simplifying the Denominator

Extracting the Real Part

Equating Real Part to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1/5

Solving the Trigonometric Equation

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Setting Up the Definite Integral

Evaluating the Integral

Final Conclusion

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Art of Simplification

The Rescue Mission

The Trigonometric Bridge

The Calculus Finale

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of all possible values of θ∈[−π,2π], for which 1−2icosθ1+icosθ​ is purely imaginary, is equal

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={θ∈(−π/2,π):1−2isinθ3+2isinθ​ is purely imaginary}. Then the sum of the elements in A is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the expression [1+2isin(2x​)sin(2x​)+cos(2x​)+itan(x)​] is real, then the set of all possible values of x is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={θ∈(0,2π):1−isinθ1+2isinθ​ is purely imaginary}. Then the sum of the elements in A is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Express 1−cosθ+2isinθ1​ in the form x+iy.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 4−icosθ3+isinθ​,θ∈[0,2π], is a real number, then the argument of sinθ+icosθ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 4−icosθ3+isinθ​, θ∈[0,2π], is a real number, then an argument of sinθ+icosθ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A value of θ for which 1−2isinθ2+3isinθ​ is purely imaginary, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of the sum ∑n=113​(in+in+1), where i=−1​, equals

Substituting Identities into $z$

Equating Real Part to $1/5$

Finding the Value of $θ$

The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i c o s θ}{1 - 2 i c o s θ}$ is purely imaginary, is equal

Let $A = {θ \in (- π /2, π) : \frac{3 + 2 i s i n θ}{1 - 2 i s i n θ} is purely imaginary}$ . Then the sum of the elements in A is :

If the expression $[\frac{sin ( \frac{x}{2} ) + cos ( \frac{x}{2} ) + i tan ( x )}{1 + 2 i sin ( \frac{x}{2} )}]$ is real, then the set of all possible values of $x$ is .........

Let $A = {θ \in (0, 2 π) : \frac{1 + 2 i s i n θ}{1 - i s i n θ} is purely imaginary}$ . Then the sum of the elements in $A$ is

Express $\frac{1}{1 - cos θ + 2 i sin θ}$ in the form $x + i y$ .

If $\frac{3 + i s i n θ}{4 - i c o s θ}, θ \in [0, 2 π]$ , is a real number, then the argument of $sin θ + i cos θ$ is :

If $\frac{3 + i s i n θ}{4 - i c o s θ}$ , $θ \in [0, 2 π]$ , is a real number, then an argument of $sin θ + i cos θ$ is:

A value of $θ$ for which $\frac{2 + 3 i s i n θ}{1 - 2 i s i n θ}$ is purely imaginary, is

The value of the sum $\sum_{n = 1}^{13} (i^{n} + i^{n + 1})$ , where $i = - 1$ , equals