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: Read the ordered pair as a complex number.
The pair $(\sin\theta,\cos\theta)$ means $z=\sin\theta+i\cos\theta$.
Step 2: Recall the inverse formula.
For $z=a+ib$, the multiplicative inverse is $z^{-1}=\dfrac{a-ib}{a^2+b^2}$.
Step 3: Plug in $a$ and $b$.
Here $a=\sin\theta$, $b=\cos\theta$, so $z^{-1}=\dfrac{\sin\theta-i\cos\theta}{\sin^2\theta+\cos^2\theta}$.
Step 4: Simplify the denominator.
Since $\sin^2\theta+\cos^2\theta=1$, we get $z^{-1}=\sin\theta-i\cos\theta$.
Step 5: Convert back to a pair.
The real part is $\sin\theta$ and the imaginary part is $-\cos\theta$, so the inverse is $(\sin\theta,-\cos\theta)$.
Step 6: Match the option.
This is option (2).
\[ \boxed{(\sin\theta,\,-\cos\theta)} \]