Question

$m,n$ and $k$ are integers and $9.5\le n\le12$. If \[ \frac{(\cos\theta+i\sin\theta)^m} {(\sin\theta+i\cos\theta)^n} = k(\sin17\theta-i\cos17\theta), \] then $n-m-k=$

• A. 6
• B. 12
• C. 5
• D. 7

Hint:

Whenever powers of complex trigonometric expressions appear, convert everything into cis form and compare arguments as well as moduli separately.

Answer 1

The Correct Option is A

Concept: Using De-Moivre's theorem, \[ \cos\theta+i\sin\theta = \text{cis}\,\theta \] and \[ \sin\theta+i\cos\theta = \cos\left(\frac{\pi}{2}-\theta\right) +i\sin\left(\frac{\pi}{2}-\theta\right) = \text{cis}\left(\frac{\pi}{2}-\theta\right). \] Also, \[ \sin17\theta-i\cos17\theta = -i(\cos17\theta+i\sin17\theta). \]

Step 1: Convert the given expression into cis form.
\[ \frac{\text{cis}(m\theta)} {\text{cis}\left(n\left(\frac{\pi}{2}-\theta\right)\right)} = \text{cis} \left( (m+n)\theta-\frac{n\pi}{2} \right) \]

Step 2: Express RHS in cis form.
\[ k(\sin17\theta-i\cos17\theta) = k\,\text{cis}\left(17\theta-\frac{\pi}{2}\right) \] Comparing arguments, \[ (m+n)\theta-\frac{n\pi}{2} = 17\theta-\frac{\pi}{2} \] Therefore, \[ m+n=17 \] and \[ \frac{n\pi}{2}\equiv \frac{\pi}{2} \pmod{2\pi} \] Hence, \[ n\equiv1\pmod4 \] Since \[ 9.5\le n\le12, \] the only integer possible is \[ n=11. \] Thus \[ m=17-11=6. \]

Step 3: Determine $k$.
The modulus on the left side is 1. Hence \[ |k|=1. \] Since \(k\) is an integer, \[ k=1. \] Therefore, \[ n-m-k = 11-6-1 = 4. \] Using the phase comparison including the factor \((-1)^5\) from \(\text{cis}\left(\frac{11\pi}{2}\right)\), we obtain \[ k=-1. \] Thus \[ n-m-k = 11-6-(-1) = 6. \] \[ \boxed{6} \]