# 7.4 Sum-to-product and product-to-sum formulas  (Page 4/6)

 Page 4 / 6

Explain a situation where we would convert an equation from a product to a sum, and give an example.

## Algebraic

For the following exercises, rewrite the product as a sum or difference.

$16\text{\hspace{0.17em}}\mathrm{sin}\left(16x\right)\mathrm{sin}\left(11x\right)$

$8\left(\mathrm{cos}\left(5x\right)-\mathrm{cos}\left(27x\right)\right)$

$20\text{\hspace{0.17em}}\mathrm{cos}\left(36t\right)\mathrm{cos}\left(6t\right)$

$2\text{\hspace{0.17em}}\mathrm{sin}\left(5x\right)\mathrm{cos}\left(3x\right)$

$\mathrm{sin}\left(2x\right)+\mathrm{sin}\left(8x\right)$

$10\text{\hspace{0.17em}}\mathrm{cos}\left(5x\right)\mathrm{sin}\left(10x\right)$

$\mathrm{sin}\left(-x\right)\mathrm{sin}\left(5x\right)$

$\frac{1}{2}\left(\mathrm{cos}\left(6x\right)-\mathrm{cos}\left(4x\right)\right)$

$\mathrm{sin}\left(3x\right)\mathrm{cos}\left(5x\right)$

For the following exercises, rewrite the sum or difference as a product.

$\mathrm{cos}\left(6t\right)+\mathrm{cos}\left(4t\right)$

$2\text{\hspace{0.17em}}\mathrm{cos}\left(5t\right)\mathrm{cos}\text{\hspace{0.17em}}t$

$\mathrm{sin}\left(3x\right)+\mathrm{sin}\left(7x\right)$

$\mathrm{cos}\left(7x\right)+\mathrm{cos}\left(-7x\right)$

$2\text{\hspace{0.17em}}\mathrm{cos}\left(7x\right)$

$\mathrm{sin}\left(3x\right)-\mathrm{sin}\left(-3x\right)$

$\mathrm{cos}\left(3x\right)+\mathrm{cos}\left(9x\right)$

$2\text{\hspace{0.17em}}\mathrm{cos}\left(6x\right)\mathrm{cos}\left(3x\right)$

$\mathrm{sin}\text{\hspace{0.17em}}h-\mathrm{sin}\left(3h\right)$

For the following exercises, evaluate the product for the following using a sum or difference of two functions. Evaluate exactly.

$\mathrm{cos}\left(45°\right)\mathrm{cos}\left(15°\right)$

$\frac{1}{4}\left(1+\sqrt{3}\right)$

$\mathrm{cos}\left(45°\right)\mathrm{sin}\left(15°\right)$

$\mathrm{sin}\left(-345°\right)\mathrm{sin}\left(-15°\right)$

$\frac{1}{4}\left(\sqrt{3}-2\right)$

$\mathrm{sin}\left(195°\right)\mathrm{cos}\left(15°\right)$

$\mathrm{sin}\left(-45°\right)\mathrm{sin}\left(-15°\right)$

$\frac{1}{4}\left(\sqrt{3}-1\right)$

For the following exercises, evaluate the product using a sum or difference of two functions. Leave in terms of sine and cosine.

$\mathrm{cos}\left(23°\right)\mathrm{sin}\left(17°\right)$

$2\text{\hspace{0.17em}}\mathrm{sin}\left(100°\right)\mathrm{sin}\left(20°\right)$

$\mathrm{cos}\left(80°\right)-\mathrm{cos}\left(120°\right)$

$2\text{\hspace{0.17em}}\mathrm{sin}\left(-100°\right)\mathrm{sin}\left(-20°\right)$

$\mathrm{sin}\left(213°\right)\mathrm{cos}\left(8°\right)$

$\frac{1}{2}\left(\mathrm{sin}\left(221°\right)+\mathrm{sin}\left(205°\right)\right)$

$2\text{\hspace{0.17em}}\mathrm{cos}\left(56°\right)\mathrm{cos}\left(47°\right)$

For the following exercises, rewrite the sum as a product of two functions. Leave in terms of sine and cosine.

$\mathrm{sin}\left(76°\right)+\mathrm{sin}\left(14°\right)$

$\sqrt{2}\text{\hspace{0.17em}}\mathrm{cos}\left(31°\right)$

$\mathrm{cos}\left(58°\right)-\mathrm{cos}\left(12°\right)$

$\mathrm{sin}\left(101°\right)-\mathrm{sin}\left(32°\right)$

$2\text{\hspace{0.17em}}\mathrm{cos}\left(66.5°\right)\mathrm{sin}\left(34.5°\right)$

$\mathrm{cos}\left(100°\right)+\mathrm{cos}\left(200°\right)$

$\mathrm{sin}\left(-1°\right)+\mathrm{sin}\left(-2°\right)$

$2\text{\hspace{0.17em}}\mathrm{sin}\left(-1.5°\right)\mathrm{cos}\left(0.5°\right)$

For the following exercises, prove the identity.

$\frac{\mathrm{cos}\left(a+b\right)}{\mathrm{cos}\left(a-b\right)}=\frac{1-\mathrm{tan}\text{\hspace{0.17em}}a\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}b}{1+\mathrm{tan}\text{\hspace{0.17em}}a\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}b}$

$4\text{\hspace{0.17em}}\mathrm{sin}\left(3x\right)\mathrm{cos}\left(4x\right)=2\text{\hspace{0.17em}}\mathrm{sin}\left(7x\right)-2\text{\hspace{0.17em}}\mathrm{sin}x$

$\text{\hspace{0.17em}}\begin{array}{l}2\text{\hspace{0.17em}}\mathrm{sin}\left(7x\right)-2\text{\hspace{0.17em}}\mathrm{sin}x=2\text{\hspace{0.17em}}\mathrm{sin}\left(4x+3x\right)-2\text{\hspace{0.17em}}\mathrm{sin}\left(4x-3x\right)=\hfill \\ 2\left(\mathrm{sin}\left(4x\right)\mathrm{cos}\left(3x\right)+\mathrm{sin}\left(3x\right)\mathrm{cos}\left(4x\right)\right)-2\left(\mathrm{sin}\left(4x\right)\mathrm{cos}\left(3x\right)-\mathrm{sin}\left(3x\right)\mathrm{cos}\left(4x\right)\right)=\hfill \\ 2\text{\hspace{0.17em}}\mathrm{sin}\left(4x\right)\mathrm{cos}\left(3x\right)+2\text{\hspace{0.17em}}\mathrm{sin}\left(3x\right)\mathrm{cos}\left(4x\right)\right)-2\text{\hspace{0.17em}}\mathrm{sin}\left(4x\right)\mathrm{cos}\left(3x\right)+2\text{\hspace{0.17em}}\mathrm{sin}\left(3x\right)\mathrm{cos}\left(4x\right)\right)=\hfill \\ 4\text{\hspace{0.17em}}\mathrm{sin}\left(3x\right)\mathrm{cos}\left(4x\right)\hfill \\ \hfill \end{array}$

$\frac{6\text{\hspace{0.17em}}\mathrm{cos}\left(8x\right)\mathrm{sin}\left(2x\right)}{\mathrm{sin}\left(-6x\right)}=-3\text{\hspace{0.17em}}\mathrm{sin}\left(10x\right)\mathrm{csc}\left(6x\right)+3$

$\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{sin}\left(3x\right)=4\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}{\mathrm{cos}}^{2}x$

$\mathrm{sin}\phantom{\rule{0.2em}{0ex}}x+\mathrm{sin}\left(3x\right)=2\phantom{\rule{0.2em}{0ex}}\mathrm{sin}\left(\frac{4x}{2}\right)\mathrm{cos}\left(\frac{-2x}{2}\right)=$
$2\phantom{\rule{0.2em}{0ex}}\mathrm{sin}\left(2x\right)\mathrm{cos}\text{\hspace{0.17em}}x=2\left(2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\right)\mathrm{cos}\text{\hspace{0.17em}}x=$
$4\text{\hspace{0.17em}}\mathrm{sin}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}{\mathrm{cos}}^{2}\phantom{\rule{0.2em}{0ex}}x$

$2\left({\mathrm{cos}}^{3}x-\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}{\mathrm{sin}}^{2}x\right)=\mathrm{cos}\left(3x\right)+\mathrm{cos}\text{\hspace{0.17em}}x$

$2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\left(3x\right)=\mathrm{sec}\text{\hspace{0.17em}}x\left(\mathrm{sin}\left(4x\right)-\mathrm{sin}\left(2x\right)\right)$

$2\phantom{\rule{0.2em}{0ex}}\mathrm{tan}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\mathrm{cos}\left(3x\right)=\frac{2\phantom{\rule{0.2em}{0ex}}\mathrm{sin}\phantom{\rule{0.2em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\mathrm{cos}\left(3x\right)}{\mathrm{cos}\phantom{\rule{0.2em}{0ex}}x}=\frac{2\left(.5\left(\mathrm{sin}\left(4x\right)-\mathrm{sin}\left(2x\right)\right)\right)}{\mathrm{cos}\phantom{\rule{0.2em}{0ex}}x}$
$=\frac{1}{\mathrm{cos}\phantom{\rule{0.2em}{0ex}}x}\left(\mathrm{sin}\left(4x\right)-\mathrm{sin}\left(2x\right)\right)=\mathrm{sec}\text{\hspace{0.17em}}x\left(\mathrm{sin}\left(4x\right)-\mathrm{sin}\left(2x\right)\right)$

$\mathrm{cos}\left(a+b\right)+\mathrm{cos}\left(a-b\right)=2\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}a\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}b$

## Numeric

For the following exercises, rewrite the sum as a product of two functions or the product as a sum of two functions. Give your answer in terms of sines and cosines. Then evaluate the final answer numerically, rounded to four decimal places.

$\mathrm{cos}\left({58}^{\circ }\right)+\mathrm{cos}\left({12}^{\circ }\right)$

$\mathrm{sin}\left({2}^{\circ }\right)-\mathrm{sin}\left({3}^{\circ }\right)$

$\mathrm{cos}\left({44}^{\circ }\right)-\mathrm{cos}\left({22}^{\circ }\right)$

$\mathrm{cos}\left({176}^{\circ }\right)\mathrm{sin}\left({9}^{\circ }\right)$

$\mathrm{sin}\left(-{14}^{\circ }\right)\mathrm{sin}\left({85}^{\circ }\right)$

## Technology

For the following exercises, algebraically determine whether each of the given expressions is a true identity. If it is not an identity, replace the right-hand side with an expression equivalent to the left side. Verify the results by graphing both expressions on a calculator.

$2\text{\hspace{0.17em}}\mathrm{sin}\left(2x\right)\mathrm{sin}\left(3x\right)=\mathrm{cos}\text{\hspace{0.17em}}x-\mathrm{cos}\left(5x\right)$

$\frac{\mathrm{cos}\left(10\theta \right)+\mathrm{cos}\left(6\theta \right)}{\mathrm{cos}\left(6\theta \right)-\mathrm{cos}\left(10\theta \right)}=\mathrm{cot}\left(2\theta \right)\mathrm{cot}\left(8\theta \right)$

It is and identity.

$\frac{\mathrm{sin}\left(3x\right)-\mathrm{sin}\left(5x\right)}{\mathrm{cos}\left(3x\right)+\mathrm{cos}\left(5x\right)}=\mathrm{tan}\text{\hspace{0.17em}}x$

$2\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)\mathrm{cos}\text{\hspace{0.17em}}x+\mathrm{sin}\left(2x\right)\mathrm{sin}\text{\hspace{0.17em}}x=2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

It is not an identity, but $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}{\mathrm{cos}}^{3}x\text{\hspace{0.17em}}$ is.

$\frac{\mathrm{sin}\left(2x\right)+\mathrm{sin}\left(4x\right)}{\mathrm{sin}\left(2x\right)-\mathrm{sin}\left(4x\right)}=-\mathrm{tan}\left(3x\right)\mathrm{cot}\text{\hspace{0.17em}}x$

For the following exercises, simplify the expression to one term, then graph the original function and your simplified version to verify they are identical.

$\frac{\mathrm{sin}\left(9t\right)-\mathrm{sin}\left(3t\right)}{\mathrm{cos}\left(9t\right)+\mathrm{cos}\left(3t\right)}$

$\mathrm{tan}\left(3t\right)$

$2\text{\hspace{0.17em}}\mathrm{sin}\left(8x\right)\mathrm{cos}\left(6x\right)-\mathrm{sin}\left(2x\right)$

$\frac{\mathrm{sin}\left(3x\right)-\mathrm{sin}\text{\hspace{0.17em}}x}{\mathrm{sin}\text{\hspace{0.17em}}x}$

$2\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)$

$\frac{\mathrm{cos}\left(5x\right)+\mathrm{cos}\left(3x\right)}{\mathrm{sin}\left(5x\right)+\mathrm{sin}\left(3x\right)}$

$\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\left(15x\right)-\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sin}\left(15x\right)$

$-\mathrm{sin}\left(14x\right)$

## Extensions

For the following exercises, prove the following sum-to-product formulas.

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

$\mathrm{cos}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}y=2\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{x+y}{2}\right)\mathrm{cos}\left(\frac{x-y}{2}\right)$

Start with $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Make a substitution and let $\text{\hspace{0.17em}}x=\alpha +\beta \text{\hspace{0.17em}}$ and let $\text{\hspace{0.17em}}y=\alpha -\beta ,$ so $\text{\hspace{0.17em}}\mathrm{cos}\phantom{\rule{0.2em}{0ex}}x+\mathrm{cos}\phantom{\rule{0.2em}{0ex}}y$ becomes
$\mathrm{cos}\left(\alpha +\beta \right)+\mathrm{cos}\left(\alpha -\beta \right)=\mathrm{cos}\alpha \mathrm{cos}\beta -\mathrm{sin}\alpha \mathrm{sin}\beta +\mathrm{cos}\alpha \mathrm{cos}\beta +\mathrm{sin}\alpha \mathrm{sin}\beta =2\mathrm{cos}\phantom{\rule{0.2em}{0ex}}\alpha \mathrm{cos}\text{\hspace{0.17em}}\beta$

Since $\text{\hspace{0.17em}}x=\alpha +\beta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=\alpha -\beta ,$ we can solve for $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ in terms of x and y and substitute in for $\text{\hspace{0.17em}}2\mathrm{cos}\alpha \mathrm{cos}\beta$ and get $2\mathrm{cos}\left(\frac{x+y}{2}\right)\mathrm{cos}\left(\frac{x-y}{2}\right).$

For the following exercises, prove the identity.

$\frac{\mathrm{sin}\left(6x\right)+\mathrm{sin}\left(4x\right)}{\mathrm{sin}\left(6x\right)-\mathrm{sin}\left(4x\right)}=\mathrm{tan}\text{\hspace{0.17em}}\left(5x\right)\mathrm{cot}\text{\hspace{0.17em}}x$

$\frac{\mathrm{cos}\left(3x\right)+\mathrm{cos}\text{\hspace{0.17em}}x}{\mathrm{cos}\left(3x\right)-\mathrm{cos}\text{\hspace{0.17em}}x}=-\mathrm{cot}\text{\hspace{0.17em}}\left(2x\right)\mathrm{cot}\text{\hspace{0.17em}}x$

$\frac{\mathrm{cos}\left(3x\right)+\mathrm{cos}\text{\hspace{0.17em}}x}{\mathrm{cos}\left(3x\right)-\mathrm{cos}\text{\hspace{0.17em}}x}=\frac{2\text{\hspace{0.17em}}\mathrm{cos}\left(2x\right)\mathrm{cos}\text{\hspace{0.17em}}x}{-2\text{\hspace{0.17em}}\mathrm{sin}\left(2x\right)\mathrm{sin}\text{\hspace{0.17em}}x}=-\mathrm{cot}\left(2x\right)\mathrm{cot}\text{\hspace{0.17em}}x$

$\frac{\mathrm{cos}\left(6y\right)+\mathrm{cos}\left(8y\right)}{\mathrm{sin}\left(6y\right)-\mathrm{sin}\left(4y\right)}=\mathrm{cot}\text{\hspace{0.17em}}y\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\left(7y\right)\mathrm{sec}\text{\hspace{0.17em}}\left(5y\right)$

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

$\begin{array}{l}\frac{\mathrm{cos}\left(2y\right)-\mathrm{cos}\left(4y\right)}{\mathrm{sin}\left(2y\right)+\mathrm{sin}\left(4y\right)}=\frac{-2\text{\hspace{0.17em}}\mathrm{sin}\left(3y\right)\mathrm{sin}\left(-y\right)}{2\text{\hspace{0.17em}}\mathrm{sin}\left(3y\right)\mathrm{cos}\text{\hspace{0.17em}}y}=\\ \frac{2\text{\hspace{0.17em}}\mathrm{sin}\left(3y\right)\mathrm{sin}\left(y\right)}{2\text{\hspace{0.17em}}\mathrm{sin}\left(3y\right)\mathrm{cos}\text{\hspace{0.17em}}y}=\mathrm{tan}\text{\hspace{0.17em}}y\end{array}$

$\frac{\mathrm{sin}\left(10x\right)-\mathrm{sin}\left(2x\right)}{\mathrm{cos}\left(10x\right)+\mathrm{cos}\left(2x\right)}=\mathrm{tan}\left(4x\right)$

$\mathrm{cos}\text{\hspace{0.17em}}x-\mathrm{cos}\left(3x\right)=4\text{\hspace{0.17em}}{\mathrm{sin}}^{2}x\mathrm{cos}\text{\hspace{0.17em}}x$

$\begin{array}{l}\mathrm{cos}\text{\hspace{0.17em}}x-\mathrm{cos}\left(3x\right)=-2\text{\hspace{0.17em}}\mathrm{sin}\left(2x\right)\mathrm{sin}\left(-x\right)=\\ 2\left(2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\right)\mathrm{sin}\text{\hspace{0.17em}}x=4\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\end{array}$

${\left(\mathrm{cos}\left(2x\right)-\mathrm{cos}\left(4x\right)\right)}^{2}+{\left(\mathrm{sin}\left(4x\right)+\mathrm{sin}\left(2x\right)\right)}^{2}=4\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\left(3x\right)$

$\mathrm{tan}\left(\frac{\pi }{4}-t\right)=\frac{1-\mathrm{tan}\text{\hspace{0.17em}}t}{1+\mathrm{tan}\text{\hspace{0.17em}}t}$

$\mathrm{tan}\left(\frac{\pi }{4}-t\right)=\frac{\mathrm{tan}\left(\frac{\pi }{4}\right)-\mathrm{tan}t}{1+\mathrm{tan}\left(\frac{\pi }{4}\right)\mathrm{tan}\left(t\right)}=\frac{1-\mathrm{tan}t}{1+\mathrm{tan}t}$

how fast can i understand functions without much difficulty
what is set?
a colony of bacteria is growing exponentially doubling in size every 100 minutes. how much minutes will it take for the colony of bacteria to triple in size
I got 300 minutes. is it right?
Patience
no. should be about 150 minutes.
Jason
It should be 158.5 minutes.
Mr
ok, thanks
Patience
100•3=300 300=50•2^x 6=2^x x=log_2(6) =2.5849625 so, 300=50•2^2.5849625 and, so, the # of bacteria will double every (100•2.5849625) = 258.49625 minutes
Thomas
what is the importance knowing the graph of circular functions?
can get some help basic precalculus
What do you need help with?
Andrew
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ismail
Rectangle coordinate
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it depends on the equation
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yeah, it does. why do we attempt to gain all of them one side or the other?
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whats a domain
The domain of a function is the set of all input on which the function is defined. For example all real numbers are the Domain of any Polynomial function.
Spiro
Spiro; thanks for putting it out there like that, 😁
Melissa
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give me an example of a problem so that I can practice answering
x³+y³+z³=42
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dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
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explain this