Can I factorize $\sin\theta_1\cos\theta_1 +\sin\theta_2\cos\theta_2$ into $ \sin(\theta_1 +\theta_2)\cos(\theta_1-\theta_2)$?

Question

I was studying Fresnel coefficients, and to get the proper answer, I need to know how to factorize $$\sin\theta_1\cos\theta_1 +\sin\theta_2\cos\theta_2$$ into $$ \sin(\theta_1 +\theta_2)\cos(\theta_1-\theta_2)$$ but I can't find an easy way to do so. Can someone help me?

Answer 1

$$T = \sin\theta_1\cos\theta_1 +\sin\theta_2\cos\theta_2 = \frac{\sin(2\theta_1)+\sin(2\theta_2)}{2}.$$ Let $x = \theta_1+\theta_2$ and $y = \theta_1-\theta_2$. Then $$T = \frac{\sin(x+y)+\sin(x-y)}{2}=\frac{2\sin(x)\cos(y)}{2}=\sin(\theta_1+\theta_2)\cos(\theta_1-\theta_2).$$

Answer 2

The following formulae are usually proved in an elementary trig class: \begin{align*} \sin(\alpha\pm\beta)&=\sin\alpha\cos\beta\pm\sin\beta\cos\alpha\\ \cos(\alpha\pm\beta)&=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta\\ \end{align*} Thus, applying these to your problem gives \begin{align*} \sin(\theta_1+\theta_2)\cos(\theta_1-\theta_2)&=(\sin\theta_1\cos\theta_2+\sin\theta_2\cos\theta_1)(\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2)\\ &=\sin\theta_1\cos\theta_1\cos^2\theta_2+\sin^2\theta_1\sin\theta_2\cos\theta_2\\ &\quad+\cos^2\theta_1\sin\theta_2\cos\theta_2+\cos\theta_1\sin\theta_1\sin^2\theta_2\\ &=\sin\theta_1\cos\theta_1+\sin\theta_2\cos\theta_2 \end{align*} after using $\sin^2\alpha+\cos^2\alpha=1$.