Product-to-Sum Trigonometric Formulas

Product-to-sum trigonometric formulas can be very helpful in simplifying a trigonometric expression by taking the product term $($such as $\sin A \sin B, \sin A \cos B,$ or $\cos A \cos B)$ and converting it into a sum.

The complete list of product-to-sum trigonometric formulas are as follows:

$$\begin{aligned} \cos A \cos B & = \frac{1}{2} \big(\cos (A - B) + \cos(A + B) \big) \\ \sin A \cos B &= \frac{1}{2} \big(\sin (A - B) + \sin(A + B) \big) \\ \cos A \sin B &= \frac{1}{2} \big(\sin (A + B) - \sin(A - B) \big)\\ \sin A \sin B &= \frac{1}{2} \big(\cos (A - B) - \cos(A + B) \big). \end{aligned}$$

The product-to-sum formulas can be obtained by observing that the sum and difference formulas for sine and cosine look very similar except for opposite signs in the middle. Then by combining the expressions, we can cancel terms.

We have

$$\begin{aligned} \cos(A + B) + \cos(A-B) &= (\cos A \cos B - \sin A \sin B) + (\cos A \cos B + \sin A \sin B)\\ &= 2 \cos A \cos B \\ \Rightarrow \frac{1}{2} \big( \cos(A + B) + \cos(A-B) \big) &= \cos A \cos B. \end{aligned}$$

The right-hand side is a product of cosine terms, while the left-hand side is a sum of cosine terms. $_\square$

Using $\cos(a+b)$ and $\cos(a-b)$ identities, we have

$$\begin{aligned} \frac{1}{2} \big(\cos(a-b) - \cos(a+b)\big) &= \frac{1}{2} \big((\cos a \cos b + \sin a \sin b) - (\cos a \cos b - \sin a \sin b)\big) \\ &= \frac{1}{2} (2 \sin a \sin b) = \sin a \sin b. \end{aligned}$$

Therefore, $\sin a \sin b=\frac{1}{2} \big(\cos(a-b) - \cos(a+b)\big).\ _\square$

Evaluate $\sin(75^\circ) \cos(15^\circ)$.

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We know that $\sin A\cos B=\frac{1}{2}\big(\sin(A-B)+\sin(A+B)\big).$ Substituting $A=75^\circ$ and $B=15^\circ$, we get

$$\begin{aligned} \sin(75^\circ) \cos(15^\circ) &=\frac{1}{2}\big(\sin(75^\circ-15^\circ)+\sin(75^\circ+15^\circ)\big)\\ &=\frac{1}{2}(\sin 60^\circ +\sin 90^\circ) \\ &=\frac{1}{2}\left(\frac{\sqrt{3}}{2}+1\right).\ _\square \end{aligned}$$

Express $\sin(x) \sin(2x) \cos(3x)$ as a sum of basic trigonometric functions (the solution should not include any products of trigonometric functions).

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The product-to-sum formulas can be helpful in solving integration problems involving the product of trigonometric ratios.

Integrate $\int \! \sin 3x \cos 4x \, \mathrm{d}x.$

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This problem may seem tough at first, but after using the product-to-sum trigonometric formula, this integral very quickly changes into a standard form.

Using the first formula stated above, the integral is equivalent to

\[\begin{align}
\phantom{=} \int \frac{1}{2} \big( \sin (3x - 4x) + \sin(3x + 4x) \big) \, \mathrm{d}x & = \int \frac{1}{2} \big( \sin (-x) + \sin(7x) \big) \, \mathrm{d}x\\ & = \frac{1}{2} \left( \cos x - \frac{\cos 7x}{7}\right)+C, \end{align}\]

where $C$ is the constant of integration. $_\square$