9.4 Sum-to-product and product-to-sum formulas  (Page 2/6)

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Then, we divide by 2 to isolate the product of sines:

$\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta =\frac{1}{2}\left[\mathrm{cos}\left(\alpha -\beta \right)-\mathrm{cos}\left(\alpha +\beta \right)\right]$

Similarly we could express the product of cosines in terms of sine or derive other product-to-sum formulas.

The product-to-sum formulas

The product-to-sum formulas are as follows:

$\begin{array}{ccc}\hfill \phantom{\rule{0.45em}{0ex}}\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta & =& \frac{1}{2}\left[\mathrm{cos}\left(\alpha -\beta \right)+\mathrm{cos}\left(\alpha +\beta \right)\right]\hfill \end{array}$
$\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta & =& \frac{1}{2}\left[\mathrm{sin}\left(\alpha +\beta \right)+\mathrm{sin}\left(\alpha -\beta \right)\right]\hfill \end{array}$
$\begin{array}{ccc}\hfill \phantom{\rule{0.5em}{0ex}}\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta & =& \frac{1}{2}\left[\mathrm{cos}\left(\alpha -\beta \right)-\mathrm{cos}\left(\alpha +\beta \right)\right]\hfill \end{array}$
$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta & =& \frac{1}{2}\left[\mathrm{sin}\left(\alpha +\beta \right)-\mathrm{sin}\left(\alpha -\beta \right)\right]\hfill \end{array}$

Express the product as a sum or difference

Write $\text{\hspace{0.17em}}\mathrm{cos}\left(3\theta \right)\text{\hspace{0.17em}}\mathrm{cos}\left(5\theta \right)\text{\hspace{0.17em}}$ as a sum or difference.

We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify.

Use the product-to-sum formula to evaluate $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{11\pi }{12}\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{12}.$

$\frac{-2-\sqrt{3}}{4}$

Expressing sums as products

Some problems require the reverse of the process we just used. The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine . Let $\text{\hspace{0.17em}}\frac{u+v}{2}=\alpha \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\frac{u-v}{2}=\beta .$

Then,

$\begin{array}{ccc}\hfill \alpha +\beta & =& \frac{u+v}{2}+\frac{u-v}{2}\hfill \\ & =& \frac{2u}{2}\hfill \\ & =& u\hfill \\ \phantom{\rule{3em}{0ex}}\hfill \alpha -\beta & =& \frac{u+v}{2}-\frac{u-v}{2}\hfill \\ & =& \frac{2v}{2}\hfill \\ & =& v\hfill \end{array}$

Thus, replacing $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ in the product-to-sum formula with the substitute expressions, we have

The other sum-to-product identities are derived similarly.

Sum-to-product formulas

The sum-to-product formulas are as follows:

$\begin{array}{ccc}\hfill \phantom{\rule{0.4em}{0ex}}\mathrm{sin}\text{\hspace{0.17em}}\alpha +\mathrm{sin}\text{\hspace{0.17em}}\beta & =& 2\mathrm{sin}\left(\frac{\alpha +\beta }{2}\right)\mathrm{cos}\left(\frac{\alpha -\beta }{2}\right)\hfill \end{array}$
$\begin{array}{ccc}\hfill \phantom{\rule{0.3em}{0ex}}\mathrm{sin}\text{\hspace{0.17em}}\alpha -\mathrm{sin}\text{\hspace{0.17em}}\beta & =& 2\mathrm{sin}\left(\frac{\alpha -\beta }{2}\right)\mathrm{cos}\left(\frac{\alpha +\beta }{2}\right)\hfill \end{array}$
$\begin{array}{ccc}\hfill \phantom{\rule{0.4em}{0ex}}\mathrm{cos}\text{\hspace{0.17em}}\alpha -\mathrm{cos}\text{\hspace{0.17em}}\beta & =& -2\mathrm{sin}\left(\frac{\alpha +\beta }{2}\right)\mathrm{sin}\left(\frac{\alpha -\beta }{2}\right)\hfill \end{array}$
$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}\alpha +\mathrm{cos}\text{\hspace{0.17em}}\beta & =& 2\mathrm{cos}\left(\frac{\alpha +\beta }{2}\right)\mathrm{cos}\left(\frac{\alpha -\beta }{2}\right)\hfill \end{array}$

Writing the difference of sines as a product

Write the following difference of sines expression as a product: $\text{\hspace{0.17em}}\mathrm{sin}\left(4\theta \right)-\mathrm{sin}\left(2\theta \right).$

We begin by writing the formula for the difference of sines.

$\mathrm{sin}\text{\hspace{0.17em}}\alpha -\mathrm{sin}\text{\hspace{0.17em}}\beta =2\mathrm{sin}\left(\frac{\alpha -\beta }{2}\right)\mathrm{cos}\left(\frac{\alpha +\beta }{2}\right)$

Substitute the values into the formula, and simplify.

$\begin{array}{ccc}\hfill \mathrm{sin}\left(4\theta \right)-\mathrm{sin}\left(2\theta \right)& =& 2\mathrm{sin}\left(\frac{4\theta -2\theta }{2}\right)\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{4\theta +2\theta }{2}\right)\hfill \\ & =& 2\mathrm{sin}\left(\frac{2\theta }{2}\right)\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{6\theta }{2}\right)\hfill \\ & =& 2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\left(3\theta \right)\hfill \end{array}$

Use the sum-to-product formula to write the sum as a product: $\text{\hspace{0.17em}}\mathrm{sin}\left(3\theta \right)+\mathrm{sin}\left(\theta \right).$

$2\mathrm{sin}\left(2\theta \right)\mathrm{cos}\left(\theta \right)$

Evaluating using the sum-to-product formula

Evaluate $\text{\hspace{0.17em}}\mathrm{cos}\left(15°\right)-\mathrm{cos}\left(75°\right).\text{\hspace{0.17em}}$ Check the answer with a graphing calculator.

We begin by writing the formula for the difference of cosines.

$\mathrm{cos}\text{\hspace{0.17em}}\alpha -\mathrm{cos}\text{\hspace{0.17em}}\beta =-2\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\alpha +\beta }{2}\right)\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\alpha -\beta }{2}\right)$

Then we substitute the given angles and simplify.

$\begin{array}{ccc}\hfill \mathrm{cos}\left(15°\right)-\mathrm{cos}\left(75°\right)& =& -2\mathrm{sin}\left(\frac{15°+75°}{2}\right)\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{15°-75°}{2}\right)\hfill \\ & =& -2\mathrm{sin}\left(45°\right)\text{\hspace{0.17em}}\mathrm{sin}\left(-30°\right)\hfill \\ & =& -2\left(\frac{\sqrt{2}}{2}\right)\left(-\frac{1}{2}\right)\hfill \\ & =& \frac{\sqrt{2}}{2}\hfill \end{array}$

Proving an identity

Prove the identity:

$\frac{\mathrm{cos}\left(4t\right)-\mathrm{cos}\left(2t\right)}{\mathrm{sin}\left(4t\right)+\mathrm{sin}\left(2t\right)}=-\mathrm{tan}\text{\hspace{0.17em}}t$

We will start with the left side, the more complicated side of the equation, and rewrite the expression until it matches the right side.

$\begin{array}{ccc}\hfill \frac{\mathrm{cos}\left(4t\right)-\mathrm{cos}\left(2t\right)}{\mathrm{sin}\left(4t\right)+\mathrm{sin}\left(2t\right)}& =& \frac{-2\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{4t+2t}{2}\right)\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{4t-2t}{2}\right)}{2\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{4t+2t}{2}\right)\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{4t-2t}{2}\right)}\hfill \\ & =& \frac{-2\text{\hspace{0.17em}}\mathrm{sin}\left(3t\right)\mathrm{sin}\text{\hspace{0.17em}}t}{2\text{\hspace{0.17em}}\mathrm{sin}\left(3t\right)\mathrm{cos}\text{\hspace{0.17em}}t}\hfill \\ & =& \frac{-\overline{)2}\overline{)\mathrm{sin}\left(3t\right)}\mathrm{sin}\text{\hspace{0.17em}}t}{\overline{)2}\overline{)\mathrm{sin}\left(3t\right)}\mathrm{cos}\text{\hspace{0.17em}}t}\hfill \\ & =& -\frac{\mathrm{sin}\text{\hspace{0.17em}}t}{\mathrm{cos}\text{\hspace{0.17em}}t}\hfill \\ & =& -\mathrm{tan}\text{\hspace{0.17em}}t\hfill \end{array}$

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