Trigonometry/For Enthusiasts/Transformation of products into sums

In this section, we shall see how to convert a product of two trigonometric functions into a sum or difference of two such functions, and vice versa.

Product into sum
We have already seen that
 * $$\sin(A)\cos(B)+\cos(A)\sin(B)=\sin(A+B)$$

and
 * $$\sin(A)\cos(B)-\cos(A)\sin(B)=\sin(A-B)$$

Adding these two equations and dividing both sides by 2, we get
 * $$\sin(A)\cos(B)=\frac{\sin(A+B)+\sin(A-B)}{2}$$

Subtracting the second from the first equation and dividing both sides by 2, we get
 * $$\cos(A)\sin(B)=\frac{\sin(A+B)+\sin(A-B)}{2}$$

We also know that
 * $$\cos(A)\cos(B)-\sin(A)\sin(B)=\cos(A+B)$$

and
 * $$\cos(A)\cos(B)+\sin(A)\sin(B)=\cos(A-B)$$

Adding these two equations and dividing both sides by 2, we get


 * $$\cos(A)\cos(B)=\frac{\cos(A+B)+\cos(A-B)}{2}$$

Subtracting the first from the second equation and dividing both sides by 2, we get
 * $$\sin(A)\sin(B)=\frac{\cos(A-B)-\cos(A+B)}{2}$$

Thus we can express:
 * 1) The product of a sine and cosine as the sum or difference of two sines;
 * 2) The product of two cosines as the sum of two cosines;
 * 3) The product of two sines as the difference of two sines.

Sum into product
Let $$C=A+B$$ and $$D=A-B$$. Then
 * $$A=\frac{C+D}{2}\ ;\ B=\frac{C-D}{2}$$

Substituting into the above expressions and multiplying both sides by two in each of them, we have:
 * $$\sin(C)+\sin(D)=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$$
 * $$\sin(C)-\sin(D)=2\cos\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$$
 * $$\cos(C)+\cos(D)=2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$$
 * $$\cos(C)-\cos(D)=-2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$$

Note the negative sign in the last formula.

These formulae are sometimes expressed in words, e.g.


 * cos plus cos = two cos half sum cos half diff.