গুণ থেকে যোগ এবং যোগ থেকে গুণের অভেদসমূহ

Product-to-sum and sum-to-product are particularly useful formulas, but can be a bit hard to remember.

To start, I want you to expand the result of $\sin{(a+b)} + \sin{(a-b)}$.

$\sin{(a+b)} + \sin{(a-b)} = \sin{a}\cos{b} + \sin{b}\cos{a} + \sin{a}\cos{b} - \sin{b}\cos{a} = 2\sin{a}\cos{b}$

We now have $\sin{(a+b)} + \sin{(a-b)} = 2\sin{a}\cos{b}$. Divide by $2$ to get $$\sin{a}\cos{b} = \frac{1}{2}(\sin{(a+b)} + \sin{(a-b)})$$

This is one of the product-to-sum formulas, since we turned the product of sines and cosines into a sum of them.

Now, using a similar approach, try to derive formulas for: $\sin{a}\sin{b}$, $\cos{a}\cos{b}$, and $\sin{b}\cos{a}$.

If you did everything correctly:

$$\begin{align*}\cos{(a-b)} - \cos{(a+b)} &= \cos{a}\cos{b} + \sin{a}\sin{b} - \cos{a}\cos{b} + \sin{a}\sin{b} \\ &= 2\sin{a}\sin{b} \\ \sin{a}\sin{b} &= \frac{1}{2}(\cos{(a-b)} - \cos{(a+b)}) \end{align*}$$ $$\begin{align*} \cos{(a+b)} + \cos{(a-b)} &= \cos{a}\cos{b} + \sin{a}\sin{b} + \cos{a}\cos{b} - \sin{a}\sin{b} \\ &= 2\cos{a}\cos{b} \\ \cos{a}\cos{b} &= \frac{1}{2}(\cos{(a+b)} + \cos{(a-b)}) \end{align*}$$ $$\begin{align*} \sin{(a+b)} - \sin{(a-b)} &= \sin{a}\cos{b} + \sin{b}\cos{a} - \sin{a}\cos{b} + \sin{b}\cos{a} \\ &= 2\sin{b}\cos{a} \\ \sin{b}\cos{a} &= \frac{1}{2}(\sin{(a+b)} - \sin{(a-b)}) \end{align*}$$

Good, now we're finished with product to sum. Let's move onto sum-to-product!

We know that by product-to-sum, $\frac{1}{2}(\sin{(x+y)} + \sin{(x-y)}) = \sin{x}\cos{y}$. Look what happens when we let $a = x+y$ and $b = x-y$. Then we have $x = \frac{a+b}{2}$ and $y = \frac{a-b}{2}$.

Substitute: $$\begin{align*} \frac{1}{2}(\sin{a} + \sin{b}) &= \sin{\left(\frac{a+b}{2}\right)}\cos{\left(\frac{a-b}{2}\right)} \\ \sin{a} + \sin{b} &= 2\sin{\left(\frac{a+b}{2}\right)}\cos{\left(\frac{a-b}{2}\right)}\end{align*}$$

This is a sum-to-product formula! Take a moment to see how we derived that, and then use it to derive the formulas for $\sin{a} - \sin{b}$, $\cos{a} + \cos{b}$, and $\cos{a} - \cos{b}$.

If you did everything correctly, you should have

$$\begin{align*} \sin{a} - \sin{b} &= 2\cos{\left(\frac{a+b}{2}\right)}\sin{\left(\frac{a-b}{2}\right)} \\ \cos{a} + \cos{b} &= 2\cos{\left(\frac{a+b}{2}\right)}\cos{\left(\frac{a-b}{2}\right)} \\ \cos{a} - \cos{b} &= -2\sin{\left(\frac{a+b}{2}\right)}\sin{\left(\frac{a-b}{2}\right)} \end{align*}$$

Well done!

So, to recap,

$$\begin{align*} \sin{a}\sin{b} &= \frac{1}{2}(\cos{(a-b)} - \cos{(a+b)}) \\ \cos{a}\cos{b} &= \frac{1}{2}(\cos{(a+b)} + \cos{(a-b)}) \\ \sin{a}\cos{b} &= \frac{1}{2}(\sin{(a+b)} + \sin{(a-b)}) \\ \sin{b}\cos{a} &= \frac{1}{2}(\sin{(a+b)} - \sin{(a-b)}) \\ \sin{a} + \sin{b} &= 2\sin{\left(\frac{a+b}{2}\right)}\cos{\left(\frac{a-b}{2}\right)} \\ \sin{a} - \sin{b} &= 2\cos{\left(\frac{a+b}{2}\right)}\sin{\left(\frac{a-b}{2}\right)} \\ \cos{a} + \cos{b} &= 2\cos{\left(\frac{a+b}{2}\right)}\cos{\left(\frac{a-b}{2}\right)} \\ \cos{a} - \cos{b} &= -2\sin{\left(\frac{a+b}{2}\right)}\sin{\left(\frac{a-b}{2}\right)} \end{align*}$$

(Take note of the irregularities in the pattern for the formulas of $\sin{a}\sin{b}$ and $\cos{a} - \cos{b}$.)

I encourage you not to memorize these formulas, as they are really hard to remember. Whenever you need them, however, you can easily derive them.