Question

Multiplicative inverse of the complex number \[ (\sin\theta,\cos\theta) \] is

• A. \((+\sin\theta,+\cos\theta)\)
• B. \((\sin\theta,-\cos\theta)\)
• C. \((\cos\theta,-\sin\theta)\)
• D. \((-\cos\theta,\sin\theta)\)

Hint:

For any complex number \(a+ib\), \[ \frac{1}{a+ib}=\frac{a-ib}{a^2+b^2} \] Use conjugate multiplication to simplify the denominator.

Answer 1

The Correct Option is B

Step 1: Interpret the ordered pair as a complex number.
A complex number represented by \[ (a,b) \] means \[ a+ib \] Therefore, \[ (\sin\theta,\cos\theta) = \sin\theta+i\cos\theta \]

Step 2: Recall the formula for multiplicative inverse.
For a complex number \[ z=a+ib, \] its multiplicative inverse is \[ \frac{1}{z} = \frac{a-ib}{a^2+b^2} \] Here, \[ a=\sin\theta, \qquad b=\cos\theta \] Thus, \[ \frac{1}{z} = \frac{\sin\theta-i\cos\theta} {\sin^2\theta+\cos^2\theta} \]

Step 3: Use the trigonometric identity.
We know that \[ \sin^2\theta+\cos^2\theta=1 \] Hence, \[ \frac{1}{z} = \sin\theta-i\cos\theta \] In ordered pair form, \[ (\sin\theta,-\cos\theta) \]

Step 4: Final conclusion.
Therefore, the multiplicative inverse is \[ \boxed{(\sin\theta,-\cos\theta)} \] which corresponds to option (2).