# Trig identity involving sum of cosines

\begin{align} y &= A\cos\left(\omega t - kx +\phi_1\right)+A\cos\left(\omega t + kx + \phi_2\right)\\[6pt] &= 2A\cos\left(\omega t+\frac{\phi_1+\phi_2}{2}\right)\cdot \cos\left(kx+\frac{\phi_2-\phi_1}{2}\right) \end{align}

Could someone please provide a detailed derivation for this? I was beaten by the arithmetic involved after trying several times without success. I tried to derive from the second line back to the first line, even this I cannot succeed... So I wonder is there some general formula that would enable me to proceed from the first line (instead of second line) directly to the second line?

• $A$ has disappeared. – Yves Daoust Jul 28 '15 at 9:22
• – lab bhattacharjee Jul 28 '15 at 9:23
• Yes this is a typo from the original book..I cannot fix it unfortunately – Rescy_ Jul 28 '15 at 9:33

## 2 Answers

$$\cos(a)= \cos\left(\frac{a+b}2\color{green}+\frac{a-b}2\right)= \cos\left(\frac{a+b}2\right)\cos\left(\frac{a-b}2\right)\color{green}-\sin\left(\frac{a+b}2\right)\sin\left(\frac{a-b}2\right),\\ \cos(b)= \cos\left(\frac{a+b}2\color{green}-\frac{a-b}2\right)= \cos\left(\frac{a+b}2\right)\cos\left(\frac{a-b}2\right)\color{green}+\sin\left(\frac{a+b}2\right)\sin\left(\frac{a-b}2\right).$$

Hence $$\cos(a)+\cos(b)=2\cos\left(\frac{a+b}2\right)\cos\left(\frac{a-b}2\right).$$

As @Yves has said, $A$ has disappeared.

However that is the sum-to-product identity for the $\cos(\cdot)$ function.

Hint for a proof: let's write $\cos(\alpha)+\cos(\beta)$ as $\cos(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2})+\cos(\frac{\beta+\alpha}{2}+\frac{\beta-\alpha}{2})$ and then use sum identity for the $\cos(\cdot)$ function. The thesis then comes from easy computations.