# How to use Eulers formula to derive the sum of angles identities?? [closed]

Okay the question is this:

Using Euler's formula, show that: $\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi$ and $\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi$.

I know Euler's formula is $e^{iθ}=\cos θ+i \sin θ$

But what now? How do I answer it?

Hint:$$e^{i(θ+\phi)}=\cos(θ+\phi)+i\sin(θ+\phi)\\e^{i(θ+\phi)}=e^{iθ}e^{i\phi}=(\cos θ+i \sin θ)(\cos \phi+i \sin \phi)$$

The question asks us to justify why

$$\cos(\theta+\varphi)=\cos\theta\cos\varphi-\sin\theta\sin\varphi$$$$\sin(\theta+\varphi)=\sin\theta\cos\varphi+\cos\theta\sin\varphi$$

The first part is to ask yourself, “How can I represent either identities in terms of $e$ and some imaginary power?” Well, since we know that

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

It’s easy to see that by replacing $\theta$ with $\theta+\varphi$, the right-hand side becomes

$$e^{i(\theta+\varphi)}=\cos(\theta+\varphi)+i\sin(\theta+\varphi)$$

The real part gives the first equation and the imaginary part gives the second. However, we can also represent the left-hand side as the product of two separate exponents. More succinctly

$$e^{i(\theta+\varphi)}=e^{i\theta}\times e^{i\varphi}$$

Now use Euler’s formula for each term of $e$ on the right-hand side so that

\begin{align*}e^{i(\theta+\varphi)} & =(\cos\theta+i\sin\theta)(\cos\varphi+i\sin\varphi)\\ & =\cos\theta\cos\varphi-\sin\theta\sin\varphi+i\sin\theta\cos\varphi+i\sin\varphi\cos\theta\end{align*}

Can you complete the rest?

• I'm sorry I don't see how to finish it – Daniel Jusere Aug 12 '18 at 22:24
• Could you show me the rest? I really don't know what to do next please – Daniel Jusere Aug 12 '18 at 22:39
• @DanielJusere\begin{align*} & \cos(\theta+\varphi)+i\sin(\theta+\varphi)=e^{i(\theta+\varphi)}\\ & =\cos\theta\cos\varphi-\sin\theta\sin\varphi+i\sin\theta\cos\varphi+i\sin\varphi\cos\theta\end{align*} – Frank W. Aug 15 '18 at 16:04