Last modified on 9 June 2012, at 14:49

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 sumEdit

We have already seen that

\displaystyle \sin(A) \cos(B) + \cos(A) \sin(B) = \sin(A+B)

and

\displaystyle \sin(A) \cos(B) - \cos(A) \sin(B) = \sin(A-B).

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

\displaystyle \sin(A) \cos(B) = \frac{1}{2} (\sin(A+B) + \sin(A-B))

Subtracting the second from the first equation and dividing both sides by 2, we get

\displaystyle \cos(A) \sin(B) = \frac{1}{2} (\sin(A+B) - \sin(A-B))

We also know that

\displaystyle \cos(A) \cos(B) - \sin(A) \sin(B) = \cos(A+B)

and

\displaystyle \cos(A) \cos(B) + \sin(A) \sin(B) = \cos(A-B).

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

\displaystyle \cos(A) \cos(B) = \frac{1}{2} (\cos(A+B) + \cos(A-B))

Subtracting the first from the second equation and dividing both sides by 2, we get

\displaystyle \sin(A) \sin(B) = \frac{1}{2} (\cos(A-B) - \cos(A+B))

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 productEdit

Let C = A+B and D = A-B. Then

\displaystyle A = \frac{1}{2} (C+D); \,\,\,\, B = \frac{1}{2} (C-D)

Substituting into the above expressions and multiplying both sides by two in each of them, we have:

\displaystyle \sin(C) + \sin(D) = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})
\displaystyle \sin(C) - \sin(D) = 2 \cos(\frac{C+D}{2}) \sin(\frac{C-D}{2})
\displaystyle \cos(C) + \cos(D) = 2 \cos(\frac{C+D}{2}) \cos(\frac{C-D}{2})
\displaystyle \cos(C) - \cos(D) = -2 \sin(\frac{C+D}{2}) \sin(\frac{C-D}{2})

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.