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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1