Question

$(\sin \theta + i \cos \theta)^n$ is equal to

Answer 1

$\cos\left(\frac{n\pi}{2}-n\theta\right) + i\sin\left(\frac{n\pi}{2}-n\theta\right)$

Answer 2

1. Rewrite the Expression:
   Notice that

   $$\sin\theta + i\cos\theta = i\cos\theta + \sin\theta = i\left(\cos\theta - i\sin\theta\right).$$

   And since

   $$e^{-i\theta}=\cos\theta-i\sin\theta,$$

   we can write:

   $$\sin\theta + i\cos\theta = i\,e^{-i\theta}.$$

2. Raising to the Power n:

   $$(\sin\theta + i\cos\theta)^n = \left(i\,e^{-i\theta}\right)^n = i^n\,e^{-in\theta}.$$

3. Expressing in Trigonometric Form:
   Using Euler’s formula again,

   $$e^{-in\theta} = \cos(n\theta)- i\sin(n\theta),$$

   so that

   $$(\sin\theta + i\cos\theta)^n = i^n \left[\cos(n\theta)- i\sin(n\theta)\right].$$

   Alternatively, since $i^n = e^{i\frac{n\pi}{2}}$, we have:

   $$(\sin\theta + i\cos\theta)^n = e^{i\left(\frac{n\pi}{2} - n\theta\right)} = \cos\left(\frac{n\pi}{2}-n\theta\right) + i\sin\left(\frac{n\pi}{2}-n\theta\right).$$