Calculate the product: $(\sin\frac{\pi}{12} + i\cos\frac{\pi}{12}) (\sin\frac{\pi}{6} + i\cos\frac{\pi}{6}) (\sin\frac{\pi}{4} + i\cos\frac{\pi}{4})$

Question

Calculate the following product:

$$ \left(\sin\frac{\pi}{12} + i\cos\frac{\pi}{12}\right) \left(\sin\frac{\pi}{6} + i\cos\frac{\pi}{6}\right) \left(\sin\frac{\pi}{4} + i\cos\frac{\pi}{4}\right) $$

What threw me off was the imaginary component of $i\,cos$, otherwise I would have used De Moivre's Theorem. Am I supposed to get rid of it and then apply the same theorem, perhaps?

Answer 1

Using $\sin\theta=\cos\left(\dfrac\pi2-\theta\right)$ $$\begin{align}P&=\left(\sin\dfrac\pi{12}+i\cos\dfrac\pi{12}\right)\left(\sin\dfrac\pi6+i\cos\dfrac\pi6\right)\left(\sin\dfrac\pi4+i\cos\dfrac\pi4\right)\\&=\left(\cos\dfrac{5\pi}{12}+i\sin\dfrac{5\pi}{12}\right)\left(\cos\dfrac\pi3+i\sin\dfrac\pi3\right)\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)\\&=e^{i\left(\tfrac{5\pi}{12}+\tfrac\pi3+\tfrac\pi4\right)}\\&=e^{i\pi}\\&=-1\end{align}$$

Using the fact that $$e^{i\theta}=\cos\theta+i\sin\theta$$ Setting $\theta=\pi$ gives $$e^{i\pi}=-1$$

Answer 2

Yes, you most certainly can. One way is to note $$\sin \theta + i \cos \theta = i(-i \sin \theta + \cos \theta) = i(\cos (-\theta) + i \sin (-\theta)) = ie^{-i\theta}$$ since $$\cos (-\theta) = \cos \theta, \\ \sin (-\theta) = -\sin \theta$$ for any angle $\theta$. Therefore, the given product becomes $$i e^{-i\pi/12} \cdot i e^{-i \pi/6} \cdot i e^{-i \pi/4},$$ and finally, since $i = e^{i\pi/2}$, you can do the rest from here.

Answer 3

hint

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

The product is

$$\Large{e^{i(\frac{3\pi}{2}-\frac{\pi}{12}-\frac{\pi}{6}-\frac{\pi}{4})}=e^{i\pi}}$$

$$=\cos(\pi)+i\sin(\pi)=-1$$

Answer 4

Hint:

I would rewrite the factors as $$\sin\theta+i\cos\theta=\cos\Bigl(\frac\pi2-\theta\Bigr)+i\sin\Bigl(\frac\pi2-\theta\Bigr), $$ and use the exponential notation: $$\cos\Bigl(\frac\pi2-\theta\Bigr)+i\sin\Bigl(\frac\pi2-\theta\Bigr)=\mathrm e^{i\bigl(\tfrac{\pi}2-\theta\bigr)}$$