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sinx + i cos 2x and cos x - i sin2x are ...

sinx + i cos 2x and cos x - i sin2x are conjugate to each other for (A) x=nπ (B) x=(n+1/2)π/2 (C) x=0 (D) no value of x

A

`x = npi`

B

`x = ("n" +(1)/(2))(pi)/(2)`

C

x=0

D

No value of x

Text Solution

AI Generated Solution

The correct Answer is:
To determine for which values of \( x \) the expressions \( \sin x + i \cos 2x \) and \( \cos x - i \sin 2x \) are conjugates of each other, we can follow these steps: ### Step 1: Define the Expressions Let: \[ z = \sin x + i \cos 2x \] The conjugate of \( z \) is: \[ \overline{z} = \sin x - i \cos 2x \] ### Step 2: Set Up the Conjugate Condition We want to find when \( \overline{z} = \cos x - i \sin 2x \). This gives us two equations by equating the real and imaginary parts: 1. Real part: \[ \sin x = \cos x \] 2. Imaginary part: \[ -\cos 2x = -\sin 2x \quad \text{(which simplifies to)} \quad \cos 2x = \sin 2x \] ### Step 3: Solve the First Equation From the equation \( \sin x = \cos x \): \[ \tan x = 1 \] This implies: \[ x = \frac{\pi}{4} + n\pi \quad \text{for } n \in \mathbb{Z} \] ### Step 4: Solve the Second Equation From the equation \( \cos 2x = \sin 2x \): \[ \tan 2x = 1 \] This implies: \[ 2x = \frac{\pi}{4} + m\pi \quad \text{for } m \in \mathbb{Z} \] Dividing by 2 gives: \[ x = \frac{\pi}{8} + \frac{m\pi}{2} \] ### Step 5: Find Common Solutions Now we need to find values of \( x \) that satisfy both equations: 1. From \( x = \frac{\pi}{4} + n\pi \) 2. From \( x = \frac{\pi}{8} + \frac{m\pi}{2} \) To find common solutions, we can set: \[ \frac{\pi}{4} + n\pi = \frac{\pi}{8} + \frac{m\pi}{2} \] Multiplying through by 8 to eliminate the fractions: \[ 2\pi + 8n\pi = \pi + 4m\pi \] Rearranging gives: \[ 8n\pi - 4m\pi = -\pi \] This simplifies to: \[ 8n - 4m = -1 \] or \[ 8n + 1 = 4m \] ### Step 6: Check for Integer Solutions Since \( m \) must be an integer, \( 8n + 1 \) must be divisible by 4. This means \( 8n + 1 \equiv 0 \mod 4 \), which simplifies to \( 1 \equiv 0 \mod 4 \), indicating that there are no integer solutions for \( n \) and \( m \). ### Conclusion Thus, the two expressions are conjugate to each other for no values of \( x \). The correct answer is: **(D) no value of x**

To determine for which values of \( x \) the expressions \( \sin x + i \cos 2x \) and \( \cos x - i \sin 2x \) are conjugates of each other, we can follow these steps: ### Step 1: Define the Expressions Let: \[ z = \sin x + i \cos 2x \] The conjugate of \( z \) is: ...
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Knowledge Check

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