# 9.2 Sum and difference identities  (Page 6/6)

 Page 6 / 6

## Verbal

Explain the basis for the cofunction identities and when they apply.

The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures $\text{\hspace{0.17em}}x,$ the second angle measures $\text{\hspace{0.17em}}\frac{\pi }{2}-x.\text{\hspace{0.17em}}$ Then $\text{\hspace{0.17em}}\mathrm{sin}x=\mathrm{cos}\left(\frac{\pi }{2}-x\right).\text{\hspace{0.17em}}$ The same holds for the other cofunction identities. The key is that the angles are complementary.

Is there only one way to evaluate $\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{5\pi }{4}\right)?\text{\hspace{0.17em}}$ Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer.

Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sin}\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)=\mathrm{cos}\left(x\right).\text{\hspace{0.17em}}$ (Hint: $\text{\hspace{0.17em}}0-x=-x$ )

$\mathrm{sin}\left(-x\right)=-\mathrm{sin}x,$ so $\text{\hspace{0.17em}}\mathrm{sin}x\text{\hspace{0.17em}}$ is odd. $\text{\hspace{0.17em}}\mathrm{cos}\left(-x\right)=\mathrm{cos}\left(0-x\right)=\mathrm{cos}x,$ so $\text{\hspace{0.17em}}\mathrm{cos}x\text{\hspace{0.17em}}$ is even.

## Algebraic

For the following exercises, find the exact value.

$\mathrm{cos}\left(\frac{7\pi }{12}\right)$

$\mathrm{cos}\left(\frac{\pi }{12}\right)$

$\frac{\sqrt{2}+\sqrt{6}}{4}$

$\mathrm{sin}\left(\frac{5\pi }{12}\right)$

$\mathrm{sin}\left(\frac{11\pi }{12}\right)$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

$\mathrm{tan}\left(-\frac{\pi }{12}\right)$

$\mathrm{tan}\left(\frac{19\pi }{12}\right)$

$-2-\sqrt{3}$

For the following exercises, rewrite in terms of $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x.$

$\mathrm{sin}\left(x+\frac{11\pi }{6}\right)$

$\mathrm{sin}\left(x-\frac{3\pi }{4}\right)$

$-\frac{\sqrt{2}}{2}\mathrm{sin}x-\frac{\sqrt{2}}{2}\mathrm{cos}x$

$\mathrm{cos}\left(x-\frac{5\pi }{6}\right)$

$\mathrm{cos}\left(x+\frac{2\pi }{3}\right)$

$-\frac{1}{2}\mathrm{cos}x-\frac{\sqrt{3}}{2}\mathrm{sin}x$

For the following exercises, simplify the given expression.

$\mathrm{csc}\left(\frac{\pi }{2}-t\right)$

$\mathrm{sec}\left(\frac{\pi }{2}-\theta \right)$

$\mathrm{csc}\theta$

$\mathrm{cot}\left(\frac{\pi }{2}-x\right)$

$\mathrm{tan}\left(\frac{\pi }{2}-x\right)$

$\mathrm{cot}x$

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

$\frac{\mathrm{tan}\left(\frac{3}{2}x\right)-\mathrm{tan}\left(\frac{7}{5}x\right)}{1+\mathrm{tan}\left(\frac{3}{2}x\right)\mathrm{tan}\left(\frac{7}{5}x\right)}$

$\mathrm{tan}\left(\frac{x}{10}\right)$

For the following exercises, find the requested information.

Given that $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}a=\frac{2}{3}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}b=-\frac{1}{4},$ with $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ both in the interval $\text{\hspace{0.17em}}\left[\frac{\pi }{2},\pi \right),$ find $\text{\hspace{0.17em}}\mathrm{sin}\left(a+b\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\left(a-b\right).$

Given that $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}a=\frac{4}{5},$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}b=\frac{1}{3},$ with $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ both in the interval $\text{\hspace{0.17em}}\left[0,\frac{\pi }{2}\right),$ find $\text{\hspace{0.17em}}\mathrm{sin}\left(a-b\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\left(a+b\right).$

$\begin{array}{ccccc}\hfill \mathrm{sin}\left(a-b\right)& =& \left(\frac{4}{5}\right)\left(\frac{1}{3}\right)-\left(\frac{3}{5}\right)\left(\frac{2\sqrt{2}}{3}\right)\hfill & =& \frac{4-6\sqrt{2}}{15}\hfill \\ \hfill \mathrm{cos}\left(a+b\right)& =& \left(\frac{3}{5}\right)\left(\frac{1}{3}\right)-\left(\frac{4}{5}\right)\left(\frac{2\sqrt{2}}{3}\right)\hfill & =& \frac{3-8\sqrt{2}}{15}\hfill \end{array}$

For the following exercises, find the exact value of each expression.

$\mathrm{sin}\left({\mathrm{cos}}^{-1}\left(0\right)-{\mathrm{cos}}^{-1}\left(\frac{1}{2}\right)\right)$

$\mathrm{cos}\left({\mathrm{cos}}^{-1}\left(\frac{\sqrt{2}}{2}\right)+{\mathrm{sin}}^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)$

$\frac{\sqrt{2}-\sqrt{6}}{4}$

$\mathrm{tan}\left({\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)-{\mathrm{cos}}^{-1}\left(\frac{1}{2}\right)\right)$

## Graphical

For the following exercises, simplify the expression, and then graph both expressions as functions to verify the graphs are identical. Confirm your answer using a graphing calculator.

$\mathrm{cos}\left(\frac{\pi }{2}-x\right)$

$\mathrm{sin}x$

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

$\mathrm{tan}\left(\frac{\pi }{3}+x\right)$

$\mathrm{cot}\left(\frac{\pi }{6}-x\right)$

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

$\mathrm{tan}\left(\frac{\pi }{4}-x\right)$

$\mathrm{cot}\left(\frac{\pi }{4}+x\right)$

$\mathrm{cos}\left(\frac{7\pi }{6}+x\right)$

$\mathrm{sin}\left(\frac{\pi }{4}+x\right)$

$\frac{\mathrm{sin}x}{\sqrt{2}}+\frac{\mathrm{cos}x}{\sqrt{2}}$

$\mathrm{cos}\left(\frac{5\pi }{4}+x\right)$

For the following exercises, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think $\text{\hspace{0.17em}}2x=x+x.$ )

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

They are the same.

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

$f\left(x\right)=\mathrm{sin}\left(3x\right)\mathrm{cos}\left(6x\right),g\left(x\right)=-\mathrm{sin}\left(3x\right)\mathrm{cos}\left(6x\right)$

They are the different, try $\text{\hspace{0.17em}}g\left(x\right)=\mathrm{sin}\left(9x\right)-\mathrm{cos}\left(3x\right)\mathrm{sin}\left(6x\right).$

$f\left(x\right)=\mathrm{sin}\left(4x\right),g\left(x\right)=\mathrm{sin}\left(5x\right)\mathrm{cos}\text{\hspace{0.17em}}x-\mathrm{cos}\left(5x\right)\mathrm{sin}\text{\hspace{0.17em}}x$

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

They are the same.

$f\left(\theta \right)=\mathrm{cos}\left(2\theta \right),g\left(\theta \right)={\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta$

$f\left(\theta \right)=\mathrm{tan}\left(2\theta \right),g\left(\theta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\theta }{1+{\mathrm{tan}}^{2}\theta }$

They are the different, try $\text{\hspace{0.17em}}g\left(\theta \right)=\frac{2\text{\hspace{0.17em}}\mathrm{tan}\theta }{1-{\mathrm{tan}}^{2}\theta }.$

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

$f\left(x\right)=\mathrm{tan}\left(-x\right),g\left(x\right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}x-\mathrm{tan}\left(2x\right)}{1-\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{tan}\left(2x\right)}$

They are different, try $\text{\hspace{0.17em}}g\left(x\right)=\frac{\mathrm{tan}x-\mathrm{tan}\left(2x\right)}{1+\mathrm{tan}x\mathrm{tan}\left(2x\right)}.$

## Technology

For the following exercises, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point.

$\mathrm{sin}\left(75°\right)$

$\mathrm{sin}\left(195°\right)$

$\mathrm{cos}\left(165°\right)$

$\mathrm{cos}\left(345°\right)$

$\frac{1+\sqrt{3}}{2\sqrt{2}},$ or 0.9659

$\mathrm{tan}\left(-15°\right)$

## Extensions

For the following exercises, prove the identities provided.

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

$\begin{array}{ccc}\hfill \mathrm{tan}\left(x+\frac{\pi }{4}\right)& =& \\ \hfill \frac{\mathrm{tan}x+\mathrm{tan}\left(\frac{\pi }{4}\right)}{1-\mathrm{tan}x\mathrm{tan}\left(\frac{\pi }{4}\right)}& =& \\ \hfill \frac{\mathrm{tan}x+1}{1-\mathrm{tan}x\left(1\right)}& =& \frac{\mathrm{tan}x+1}{1-\mathrm{tan}x}\hfill \end{array}$

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

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

$\begin{array}{ccc}\hfill \frac{\mathrm{cos}\left(a+b\right)}{\mathrm{cos}a\mathrm{cos}b}& =& \\ \hfill \frac{\mathrm{cos}a\mathrm{cos}b}{\mathrm{cos}a\mathrm{cos}b}-\frac{\mathrm{sin}a\mathrm{sin}b}{\mathrm{cos}a\mathrm{cos}b}& =& 1-\mathrm{tan}a\mathrm{tan}b\hfill \end{array}$

$\mathrm{cos}\left(x+y\right)\mathrm{cos}\left(x-y\right)={\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}y$

$\frac{\mathrm{cos}\left(x+h\right)-\mathrm{cos}\text{\hspace{0.17em}}x}{h}=\mathrm{cos}\text{\hspace{0.17em}}x\frac{\mathrm{cos}\text{\hspace{0.17em}}h-1}{h}-\mathrm{sin}\text{\hspace{0.17em}}x\frac{\mathrm{sin}\text{\hspace{0.17em}}h}{h}$

$\begin{array}{ccc}\hfill \frac{\mathrm{cos}\left(x+h\right)-\mathrm{cos}x}{h}& =& \\ \hfill \frac{\mathrm{cos}x\mathrm{cosh}-\mathrm{sin}x\mathrm{sinh}-\mathrm{cos}x}{h}& =& \\ \hfill \frac{\mathrm{cos}x\left(\mathrm{cosh}-1\right)-\mathrm{sin}x\mathrm{sinh}}{h}& =& \mathrm{cos}x\frac{\mathrm{cos}h-1}{h}-\mathrm{sin}x\frac{\mathrm{sin}h}{h}\hfill \end{array}$

For the following exercises, prove or disprove the statements.

$\mathrm{tan}\left(u+v\right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}u+\mathrm{tan}\text{\hspace{0.17em}}v}{1-\mathrm{tan}\text{\hspace{0.17em}}u\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}v}$

$\mathrm{tan}\left(u-v\right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}u-\mathrm{tan}\text{\hspace{0.17em}}v}{1+\mathrm{tan}\text{\hspace{0.17em}}u\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}v}$

True

$\frac{\mathrm{tan}\left(x+y\right)}{1+\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x}=\frac{\mathrm{tan}\text{\hspace{0.17em}}x+\mathrm{tan}\text{\hspace{0.17em}}y}{1-{\mathrm{tan}}^{2}x\text{\hspace{0.17em}}{\mathrm{tan}}^{2}y}$

If $\text{\hspace{0.17em}}\alpha ,\beta ,$ and $\text{\hspace{0.17em}}\gamma \text{\hspace{0.17em}}$ are angles in the same triangle, then prove or disprove $\text{\hspace{0.17em}}\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\text{\hspace{0.17em}}\gamma .$

True. Note that $\text{\hspace{0.17em}}\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\left(\pi -\gamma \right)\text{\hspace{0.17em}}$ and expand the right hand side.

If $\text{\hspace{0.17em}}\alpha ,\beta ,$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are angles in the same triangle, then prove or disprove $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha +\mathrm{tan}\text{\hspace{0.17em}}\beta +\mathrm{tan}\text{\hspace{0.17em}}\gamma =\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\gamma$

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
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
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
which of these functions is not uniformly continuous on 0,1
solve this equation by completing the square 3x-4x-7=0
X=7
Muustapha
=7
mantu
x=7
mantu
3x-4x-7=0 -x=7 x=-7
Kr
x=-7
mantu
9x-16x-49=0 -7x=49 -x=7 x=7
mantu
what's the formula
Modress
-x=7
Modress
new member
siame
what is trigonometry
deals with circles, angles, and triangles. Usually in the form of Soh cah toa or sine, cosine, and tangent
Thomas
solve for me this equational y=2-x
what are you solving for
Alex
solve x
Rubben
you would move everything to the other side leaving x by itself. subtract 2 and divide -1.
Nikki
then I got x=-2
Rubben
it will b -y+2=x
Alex
goodness. I'm sorry. I will let Alex take the wheel.
Nikki
ouky thanks braa
Rubben
I think he drive me safe
Rubben
how to get 8 trigonometric function of tanA=0.5, given SinA=5/13? Can you help me?m
More example of algebra and trigo
What is Indices
If one side only of a triangle is given is it possible to solve for the unkown two sides?
cool
Rubben
kya
Khushnama
please I need help in maths
Okey tell me, what's your problem is?
Navin