# 9.1 Solving trigonometric equations with identities  (Page 5/9)

 Page 5 / 9

## Simplify by rewriting and using substitution

Simplify the expression by rewriting and using identities:

${\mathrm{csc}}^{2}\theta -{\mathrm{cot}}^{2}\theta$

$1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta$

Now we can simplify by substituting $\text{\hspace{0.17em}}1+{\mathrm{cot}}^{2}\theta \text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}{\mathrm{csc}}^{2}\theta .\text{\hspace{0.17em}}$ We have

$\begin{array}{ccc}\hfill {\mathrm{csc}}^{2}\theta -{\mathrm{cot}}^{2}\theta & =& 1+{\mathrm{cot}}^{2}\theta -{\mathrm{cot}}^{2}\theta \hfill \\ & =& 1\hfill \end{array}$

Use algebraic techniques to verify the identity: $\text{\hspace{0.17em}}\frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{1+\mathrm{sin}\text{\hspace{0.17em}}\theta }=\frac{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }.$

(Hint: Multiply the numerator and denominator on the left side by $\text{\hspace{0.17em}}1-\mathrm{sin}\text{\hspace{0.17em}}\theta .\right)$

$\begin{array}{ccc}\hfill \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{1+\mathrm{sin}\text{\hspace{0.17em}}\theta }\left(\frac{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }\right)& =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta \left(1-\mathrm{sin}\text{\hspace{0.17em}}\theta \right)}{1-{\mathrm{sin}}^{2}\theta }\hfill \\ & =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta \left(1-\mathrm{sin}\text{\hspace{0.17em}}\theta \right)}{{\mathrm{cos}}^{2}\theta }\hfill \\ & =& \frac{1-\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }\hfill \end{array}$

Access these online resources for additional instruction and practice with the fundamental trigonometric identities.

## Key equations

 Pythagorean identities $\begin{array}{l}{\mathrm{cos}}^{2}\theta +{\mathrm{sin}}^{2}\theta =1\\ 1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta \\ 1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta \end{array}$ Even-odd identities $\begin{array}{ccc}\mathrm{tan}\left(-\theta \right)& =& -\mathrm{tan}\text{\hspace{0.17em}}\theta \\ \mathrm{cot}\left(-\theta \right)& =& -\mathrm{cot}\text{\hspace{0.17em}}\theta \\ \mathrm{sin}\left(-\theta \right)& =& -\mathrm{sin}\text{\hspace{0.17em}}\theta \\ \mathrm{csc}\left(-\theta \right)& =& -\mathrm{csc}\text{\hspace{0.17em}}\theta \\ \mathrm{cos}\left(-\theta \right)& =& \mathrm{cos}\text{\hspace{0.17em}}\theta \\ \mathrm{sec}\left(-\theta \right)& =& \mathrm{sec}\text{\hspace{0.17em}}\theta \end{array}$ Reciprocal identities $\begin{array}{ccc}\mathrm{sin}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{csc}\text{\hspace{0.17em}}\theta }\\ \mathrm{cos}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{sec}\text{\hspace{0.17em}}\theta }\\ \mathrm{tan}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{cot}\text{\hspace{0.17em}}\theta }\\ \mathrm{csc}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{sin}\text{\hspace{0.17em}}\theta }\\ \mathrm{sec}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{cos}\text{\hspace{0.17em}}\theta }\\ \mathrm{cot}\text{\hspace{0.17em}}\theta & =& \frac{1}{\mathrm{tan}\text{\hspace{0.17em}}\theta }\end{array}$ Quotient identities $\begin{array}{ccc}\mathrm{tan}\text{\hspace{0.17em}}\theta & =& \frac{\mathrm{sin}\text{\hspace{0.17em}}\theta }{\mathrm{cos}\text{\hspace{0.17em}}\theta }\\ \mathrm{cot}\text{\hspace{0.17em}}\theta & =& \frac{\mathrm{cos}\text{\hspace{0.17em}}\theta }{\mathrm{sin}\text{\hspace{0.17em}}\theta }\end{array}$

## Key concepts

• There are multiple ways to represent a trigonometric expression. Verifying the identities illustrates how expressions can be rewritten to simplify a problem.
• Graphing both sides of an identity will verify it. See [link] .
• Simplifying one side of the equation to equal the other side is another method for verifying an identity. See [link] and [link] .
• The approach to verifying an identity depends on the nature of the identity. It is often useful to begin on the more complex side of the equation. See [link] .
• We can create an identity and then verify it. See [link] .
• Verifying an identity may involve algebra with the fundamental identities. See [link] and [link] .
• Algebraic techniques can be used to simplify trigonometric expressions. We use algebraic techniques throughout this text, as they consist of the fundamental rules of mathematics. See [link] , [link] , and [link] .

## Verbal

We know $\text{\hspace{0.17em}}g\left(x\right)=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is an even function, and $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\left(x\right)=\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ are odd functions. What about $\text{\hspace{0.17em}}G\left(x\right)={\mathrm{cos}}^{2}x,F\left(x\right)={\mathrm{sin}}^{2}x,$ and $\text{\hspace{0.17em}}H\left(x\right)={\mathrm{tan}}^{2}x?\text{\hspace{0.17em}}$ Are they even, odd, or neither? Why?

All three functions, $\text{\hspace{0.17em}}F,G,$ and $H,$ are even.

This is because $\text{\hspace{0.17em}}F\left(-x\right)=\mathrm{sin}\left(-x\right)\mathrm{sin}\left(-x\right)=\left(-\mathrm{sin}\text{\hspace{0.17em}}x\right)\left(-\mathrm{sin}\text{\hspace{0.17em}}x\right)={\mathrm{sin}}^{2}x=F\left(x\right),G\left(-x\right)=\mathrm{cos}\left(-x\right)\mathrm{cos}\left(-x\right)=\mathrm{cos}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x={\mathrm{cos}}^{2}x=G\left(x\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}H\left(-x\right)=\mathrm{tan}\left(-x\right)\mathrm{tan}\left(-x\right)=\left(-\mathrm{tan}\text{\hspace{0.17em}}x\right)\left(-\mathrm{tan}\text{\hspace{0.17em}}x\right)={\mathrm{tan}}^{2}x=H\left(x\right).$

Examine the graph of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ on the interval $\text{\hspace{0.17em}}\left[-\pi ,\pi \right].\text{\hspace{0.17em}}$ How can we tell whether the function is even or odd by only observing the graph of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{sec}\text{\hspace{0.17em}}x?$

After examining the reciprocal identity for $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}t,$ explain why the function is undefined at certain points.

When $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t=0,$ then $\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}t=\frac{1}{0},$ which is undefined.

All of the Pythagorean identities are related. Describe how to manipulate the equations to get from $\text{\hspace{0.17em}}{\mathrm{sin}}^{2}t+{\mathrm{cos}}^{2}t=1\text{\hspace{0.17em}}$ to the other forms.

## Algebraic

For the following exercises, use the fundamental identities to fully simplify the expression.

$\mathrm{sin}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{sin}\text{\hspace{0.17em}}x$

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

$\mathrm{tan}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}{\mathrm{cos}}^{2}x$

$\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{csc}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{cot}\left(-x\right)$

$\frac{\mathrm{cot}\text{\hspace{0.17em}}t+\mathrm{tan}\text{\hspace{0.17em}}t}{\mathrm{sec}\left(-t\right)}$

$\mathrm{csc}\text{\hspace{0.17em}}t$

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

$-\mathrm{tan}\left(-x\right)\mathrm{cot}\left(-x\right)$

$-1$

$\frac{-\mathrm{sin}\left(-x\right)\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x}{\mathrm{cot}\text{\hspace{0.17em}}x}$

$\frac{1+{\mathrm{tan}}^{2}\theta }{{\mathrm{csc}}^{2}\theta }+{\mathrm{sin}}^{2}\theta +\frac{1}{{\mathrm{sec}}^{2}\theta }$

${\mathrm{sec}}^{2}x$

$\left(\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{{\mathrm{csc}}^{2}x}+\frac{\mathrm{tan}\text{\hspace{0.17em}}x}{{\mathrm{sec}}^{2}x}\right)\left(\frac{1+\mathrm{tan}\text{\hspace{0.17em}}x}{1+\mathrm{cot}\text{\hspace{0.17em}}x}\right)-\frac{1}{{\mathrm{cos}}^{2}x}$

$\frac{1-{\mathrm{cos}}^{2}\text{\hspace{0.17em}}x}{{\mathrm{tan}}^{2}\text{\hspace{0.17em}}x}+2\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}x$

${\mathrm{sin}}^{2}x+1$

For the following exercises, simplify the first trigonometric expression by writing the simplified form in terms of the second expression.

$\frac{\mathrm{tan}\text{\hspace{0.17em}}x+\mathrm{cot}\text{\hspace{0.17em}}x}{\mathrm{csc}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x$

$\frac{\mathrm{sec}\text{\hspace{0.17em}}x+\mathrm{csc}\text{\hspace{0.17em}}x}{1+\mathrm{tan}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x}$

$\frac{\mathrm{cos}\text{\hspace{0.17em}}x}{1+\mathrm{sin}\text{\hspace{0.17em}}x}+\mathrm{tan}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x}-\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x$

$\frac{1}{\mathrm{cot}\text{\hspace{0.17em}}x}$

$\frac{1}{1-\mathrm{cos}\text{\hspace{0.17em}}x}-\frac{\mathrm{cos}\text{\hspace{0.17em}}x}{1+\mathrm{cos}\text{\hspace{0.17em}}x};\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

$\left(\mathrm{sec}\text{\hspace{0.17em}}x+\mathrm{csc}\text{\hspace{0.17em}}x\right)\left(\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\right)-2-\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}x$

$\mathrm{tan}\text{\hspace{0.17em}}x$

$-4\mathrm{sec}\text{\hspace{0.17em}}x\mathrm{tan}\text{\hspace{0.17em}}x$

$\mathrm{tan}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sec}\text{\hspace{0.17em}}x$

$\mathrm{sec}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x$

$±\sqrt{\frac{1}{{\mathrm{cot}}^{2}x}+1}$

$\mathrm{sec}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x$

$\frac{±\sqrt{1-{\mathrm{sin}}^{2}x}}{\mathrm{sin}\text{\hspace{0.17em}}x}$

$\mathrm{cot}\text{\hspace{0.17em}}x;\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

For the following exercises, verify the identity.

$\mathrm{cos}\text{\hspace{0.17em}}x-{\mathrm{cos}}^{3}x=\mathrm{cos}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}x$

$\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}x-{\mathrm{cos}}^{3}x& =& \mathrm{cos}\text{\hspace{0.17em}}x\left(1-{\mathrm{cos}}^{2}x\right)\hfill \\ & =& \mathrm{cos}\text{\hspace{0.17em}}x{\mathrm{sin}}^{2}x\hfill \end{array}$

$\mathrm{cos}\text{\hspace{0.17em}}x\left(\mathrm{tan}\text{\hspace{0.17em}}x-\mathrm{sec}\left(-x\right)\right)=\mathrm{sin}\text{\hspace{0.17em}}x-1$

$\frac{1+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{1}{{\mathrm{cos}}^{2}x}+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=1+2\text{\hspace{0.17em}}{\mathrm{tan}}^{2}x$

$\frac{1+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{1}{{\mathrm{cos}}^{2}x}+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}={\mathrm{sec}}^{2}x+{\mathrm{tan}}^{2}x={\mathrm{tan}}^{2}x+1+{\mathrm{tan}}^{2}x=1+2{\mathrm{tan}}^{2}x$

${\left(\mathrm{sin}\text{\hspace{0.17em}}x+\mathrm{cos}\text{\hspace{0.17em}}x\right)}^{2}=1+2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}x\mathrm{cos}\text{\hspace{0.17em}}x$

${\mathrm{cos}}^{2}x-{\mathrm{tan}}^{2}x=2-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x$

${\mathrm{cos}}^{2}x-{\mathrm{tan}}^{2}x=1-{\mathrm{sin}}^{2}x-\left({\mathrm{sec}}^{2}x-1\right)=1-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x+1=2-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x$

## Extensions

For the following exercises, prove or disprove the identity.

$\frac{1}{1+\mathrm{cos}\text{\hspace{0.17em}}x}-\frac{1}{1-\mathrm{cos}\left(-x\right)}=-2\text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\mathrm{csc}\text{\hspace{0.17em}}x$

${\mathrm{csc}}^{2}x\left(1+{\mathrm{sin}}^{2}x\right)={\mathrm{cot}}^{2}x$

False

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

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

False

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

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

Proved with negative and Pythagorean identities

For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression.

$\frac{{\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta }{1-{\mathrm{tan}}^{2}\theta }={\mathrm{sin}}^{2}\theta$

$3\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta +4\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta =3+{\mathrm{cos}}^{2}\theta$

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

$\frac{\mathrm{sec}\text{\hspace{0.17em}}\theta +\mathrm{tan}\text{\hspace{0.17em}}\theta }{\mathrm{cot}\text{\hspace{0.17em}}\theta +\mathrm{cos}\text{\hspace{0.17em}}\theta }={\mathrm{sec}}^{2}\theta$

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
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
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
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
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
Jeffrey
Jeffrey
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
which equation
Miranda
I dont know. lol
Jeffrey
Miranda
Jeffrey
answer and questions in exercise 11.2 sums
how do u calculate inequality of irrational number?
Alaba
give me an example
Chris
and I will walk you through it
Chris
cos (-z)= cos z .
cos(- z)=cos z
Mustafa