Sine and cosine of the sum and difference of angles

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Here I will illustrate how to use the complex exponential to derive some useful trigonometric identities. Lets start with the Euler identities:

\begin{equation}e^{i\theta}=\cos\theta + i\sin\theta\end{equation}
\begin{equation}e^{-i\theta}=\cos\theta – i\sin\theta\end{equation}

Now put $\theta = A+B$ in equation (1) and use the rule for multiplication of exponentials:

\begin{equation}e^{i(A+B)}=e^{iA}e^{iB}\end{equation}

Now use equation (1) to expand both sides of equation (3):

\begin{equation}\cos(A+B)+i\sin(A+B)=\left(\cos A+i\sin A\right)\left(\cos B+i\sin B\right)\end{equation}

Expanding the product in the right side of equation (4) and taking the real and imaginary parts on both sides of the resultant equation we obtain

\begin{equation}\cos(A+B)=\cos A\cos B – \sin A\sin B\end{equation}
\begin{equation}\sin(A+B)=\sin A\cos B + \sin B\cos A\end{equation}

If we put $-B$ in place of $\ B$ in equations (5) and (6), and use the properties $\sin(-B)=-\sin B$, $\cos(-B)=\cos B$, we obtain

\begin{equation}\cos(A-B)=\cos A\cos B + \sin A\sin B\end{equation}
\begin{equation}\sin(A-B)=\sin A\cos B – \sin B\cos A\end{equation}

The sum and difference of equations (7) and (5) leads to

\begin{equation}2\cos A\cos B=\cos(A-B)+\cos(A+B)\end{equation}
\begin{equation}2\sin A\sin B=\cos(A-B)-\cos(A+B)\end{equation}

The sum of equations (6) and (8) leads to (the difference of these equations is not necessary as will be shown below)

\begin{equation}2\sin A\cos B=\sin(A+B)+\sin(A-B)\end{equation}

Common identities used in the analysis of modulation of waves and resonance are obtained from equations (9), (10) and (11) by changing variables as follows:

\begin{equation}A+B=\alpha\hspace{0.3cm};\hspace{0.3cm}A-B=\beta\end{equation}
\begin{equation}A=\frac{\alpha+\beta}{2}\hspace{0.3cm};\hspace{0.3cm}B=\frac{\alpha-\beta}{2}\end{equation}

With help of equations (12) and (13) we can rewrite equations (9), (10) and (11) as follows

\begin{equation}\cos\alpha+\cos\beta=2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)\end{equation}
\begin{equation}\cos\alpha-\cos\beta=-2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)\end{equation}
\begin{equation}\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)\end{equation}

The identity for the difference of sines is obtained from equation (16) by putting $-\beta$ in place of $\beta$:

\begin{equation}\sin\alpha-\sin\beta=2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)\end{equation}