If the expression [sin(x/2) + cos(x/2) + i tan(x)] / [1 + 2i sin(x/2)] is real, then the set of all

Q: Wait, why can't I just set the numerator to be real and ignore the denominator? Isn't the imaginary part just the $i$ term in the numerator?

That is a classic trap! If you have a complex number in the form $\frac{A+iB}{C+iD}$, the imaginary part is NOT just $B$. The denominator $C+iD$ mixes the real and imaginary parts when you divide. You must rationalize the denominator first to separate the real and imaginary components clearly. Never assume the numerator alone dictates the reality of the fraction.

Q: Do I need to worry about the denominator being zero? What if my solution makes the denominator vanish?

That is the mark of a true JEE aspirant—always checking the domain! In this problem, the denominator is $1 + 2i \sin(\frac{x}{2})$. For this to be zero, $\sin(\frac{x}{2})$ would have to be $\frac{i}{2}$. Since we are looking for real values of $x$, $\sin(\frac{x}{2})$ must be a real number between $-1$ and $1$. Thus, the denominator can never be zero for any real $x$. You are safe!

Q: Is there a faster way than expanding the entire numerator? It looks like it will take forever.

You are absolutely right to look for efficiency. You do not need to expand the entire numerator. Since you only care about the imaginary part, you only need to multiply terms that produce an $i$. Specifically, (Real part of numerator $\times$ Imaginary part of conjugate) + (Imaginary part of numerator $\times$ Real part of conjugate). This 'sniper' approach saves time and prevents messy algebraic errors.

Question

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

Sigma Percentile · Accepted Answer

### The Art of Seeing Through the Complex Fog

Welcome, student. Today, we are not just solving a problem; we are peeling back the layers of a complex expression to reveal the elegant trigonometric truth hidden beneath.

When you first look at an expression like

$$z = \frac{\sin(\frac{x}{2}) + \cos(\frac{x}{2}) + i 	an(x)}{1 + 2i \sin(\frac{x}{2})}$$

it is natural to feel a moment of hesitation. It looks cluttered and intimidating, but in the world of JEE Advanced, intimidation is just a sign that you are about to learn something profound.

### Phase 1: The Rationalization Strategy

The problem asks us to find the values of $x$ for which this expression is purely real. In the realm of complex numbers, a number $z = a + ib$ is real if and only if its imaginary part, $b$, is exactly zero.

However, our expression has an $i$ in the denominator. As long as that $i$ is lurking in the basement of our fraction, we cannot easily identify the imaginary part.

We need to perform a 'clean-up' using the complex conjugate. By multiplying the numerator and the denominator by $1 - 2i \sin(\frac{x}{2})$, we transform the denominator into a purely real number.

Because $(1 + 2i \sin(\frac{x}{2}))(1 - 2i \sin(\frac{x}{2}))$ follows the identity $(a+ib)(a-ib) = a^2 + b^2$, our denominator becomes:

$$1 + 4 \sin^2\left(\frac{x}{2}ight)$$

The $i$ is now gone from the denominator, and the path is clear.

### Phase 2: The Sniper’s Approach to Multiplication

Now, we turn our attention to the numerator. Many students will instinctively try to expand the entire product, but you should not do that. We only care about the imaginary part.

We only need to pick the terms that result in an $i$:
1. The term $i 	an(x)$ multiplied by the real $1$ gives us $i 	an(x)$.
2. The real part $(\sin(\frac{x}{2}) + \cos(\frac{x}{2}))$ multiplied by the imaginary part $-2i \sin(\frac{x}{2})$ gives us $-2i \sin(\frac{x}{2})(\sin(\frac{x}{2}) + \cos(\frac{x}{2}))$.

By focusing only on these, we have isolated the imaginary part of the numerator. Setting this imaginary part to zero gives us the equation:

$$	an(x) - 2 \sin\left(\frac{x}{2}ight)\left(\sin\left(\frac{x}{2}ight) + \cos\left(\frac{x}{2}ight)ight) = 0$$

### Phase 3: The Trigonometric Bridge

Now we have an equation that mixes half-angles and full angles. We use the identities $2 \sin^2(\frac{x}{2}) = 1 - \cos(x)$ and $2 \sin(\frac{x}{2}) \cos(\frac{x}{2}) = \sin(x)$.

Substituting these into our equation, we get:

$$	an(x) - (1 - \cos(x)) - \sin(x) = 0$$

By rewriting $	an(x)$ as $\frac{\sin(x)}{\cos(x)}$ and multiplying through by $\cos(x)$, we arrive at:

$$\sin(x) - \cos(x) + \cos^2(x) - \sin(x) \cos(x) = 0$$

### Phase 4: The Factorization Victory

Finally, we reach the endgame. We group the first two terms and factor out a $-\cos(x)$ from the last two terms:

$$(\sin(x) - \cos(x)) - \cos(x)(\sin(x) - \cos(x)) = 0$$

We can factor out the common binomial $(\sin(x) - \cos(x))$, leaving us with:

$$(1 - \cos(x))(\sin(x) - \cos(x)) = 0$$

This reduces to two simple, elegant cases:

1. $1 - \cos(x) = 0 \Rightarrow \cos(x) = 1 \Rightarrow \mathbf{x = 2n\pi}$
2. $\sin(x) - \cos(x) = 0 \Rightarrow 	an(x) = 1 \Rightarrow \mathbf{x = n\pi + \frac{\pi}{4}}$

Take a breath. You have navigated the complex plane, utilized trigonometric identities, and mastered algebraic factorization. This is the journey of simplifying the complex into the simple.

Animated Solution for Mathematics - Complex Numbers: 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 .........

Visualized Solution

Defining the Complex Number

Rationalizing the Denominator

Simplifying the Denominator

Extracting the Imaginary Part

Setting Imaginary Part to Zero

Applying Trigonometric Identities

Substituting the Identities

Converting to Sine and Cosine

Factorizing the Equation

Finding the General Solutions

The Sigma Insight: Algebraic Operations on Complex Numbers

The Art of Seeing Through the Complex Fog

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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 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

For positive integers $n_{1}, n_{2}$ , the value of the expression $(1 + i)^{n_{1}} + (1 + i^{3})^{n_{1}} + (1 + i^{5})^{n_{2}} + (1 + i^{7})^{n_{2}}$ , where $i = - 1$ is a real number if and only if

Find the real values of $x$ and $y$ for which the following equation is satisfied $\frac{( 1 + i ) x - 2 i}{3 + i} + \frac{( 2 - 3 i ) y + i}{3 - i} = i$ .

Let $a, b, x$ and $y$ be real numbers such that $a - b = 1$ and $y \neq = 0$ . If the complex number $z = x + i y$ satisfies $Im (\frac{a z + b}{z + 1}) = y$ , then which of the following is(are) possible value(s) of $x$ ?

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:

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 $(\frac{1 + i}{1 - i})^{x} = 1$ then

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 :