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

 Page 4 / 6

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

$\begin{array}{l}2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\left(3x\right)=\frac{2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\left(3x\right)}{\mathrm{cos}\text{\hspace{0.17em}}x}=\frac{2\left(.5\left(\mathrm{sin}\left(4x\right)-\mathrm{sin}\left(2x\right)\right)\right)}{\mathrm{cos}\text{\hspace{0.17em}}x}=\\ \frac{1}{\mathrm{cos}\text{\hspace{0.17em}}x}\left(\mathrm{sin}\left(4x\right)-\mathrm{sin}\left(2x\right)\right)=\mathrm{sec}\text{\hspace{0.17em}}x\left(\mathrm{sin}\left(4x\right)-\mathrm{sin}\left(2x\right)\right)\end{array}$

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

## 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.

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

$2\text{\hspace{0.17em}}\mathrm{cos}\left(35°\right)\mathrm{cos}\left(23°\right),\text{1.5081}$

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

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

$-2\text{\hspace{0.17em}}\mathrm{sin}\left(33°\right)\mathrm{sin}\left(11°\right),-0.2078$

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

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

$\frac{1}{2}\left(\mathrm{cos}\left(99°\right)-\mathrm{cos}\left(71°\right)\right),-0.2410$

## Technology

For the following exercises, algebraically determine whether each of the given equation is an 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 an 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 $\mathrm{cos}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}y$ becomes $\begin{array}{}\\ \mathrm{cos}\left(\alpha +\beta \right)+\mathrm{cos}\left(\alpha -\beta \right)=\mathrm{cos}\alpha \mathrm{cos}\beta -\mathrm{sin}\alpha \mathrm{sin}\beta +\mathrm{cos}\alpha \mathrm{cos}\beta +\mathrm{sin}\alpha \mathrm{sin}\beta =\\ 2\mathrm{cos}\phantom{\rule{0.2em}{0ex}}\alpha \mathrm{cos}\phantom{\rule{0.2em}{0ex}}\beta \end{array}$

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

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

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

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

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

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

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

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

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

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

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

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

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

By the definition, is such that 0!=1.why?
(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
hatdog
Mark
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching
bsc F. y algebra and trigonometry pepper 2
given that x= 3/5 find sin 3x
4
DB
remove any signs and collect terms of -2(8a-3b-c)
-16a+6b+2c
Will
Joeval
(x2-2x+8)-4(x2-3x+5)
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
master
Soo sorry (5±Root11* i)/3
master
Mukhtar
2x²-6x+1=0
Ife
explain and give four example of hyperbolic function
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
y2=4ax= y=4ax/2. y=2ax
akash
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
A banana.
Yaona
a function
Daniel
a function
emmanuel
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda