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

 Page 5 / 9

## Rewriting a trigonometric expression using the difference of squares

Rewrite the trigonometric expression: $\text{\hspace{0.17em}}4\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta -1.$

Notice that both the coefficient and the trigonometric expression in the first term are squared, and the square of the number 1 is 1. This is the difference of squares. Thus,

Rewrite the trigonometric expression: $\text{\hspace{0.17em}}25-9\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}\theta .$

This is a difference of squares formula: $\text{\hspace{0.17em}}25-9\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\text{\hspace{0.17em}}\theta =\left(5-3\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \right)\left(5+3\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \right).$

## Simplify by rewriting and using substitution

Simplify the expression by rewriting and using identities:

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

We can start with the Pythagorean identity.

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

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

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 $\text{\hspace{0.17em}}1-\mathrm{sin}\text{\hspace{0.17em}}\theta .\right)$

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

## Key equations

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

## 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 by simplifying an expression and then verifying 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 $\text{\hspace{0.17em}}H,$ 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.

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

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

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

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

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

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

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

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

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

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

$-1$

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

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

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

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

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

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

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.

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

Answers will vary. Sample proof:

$\mathrm{cos}\text{\hspace{0.17em}}x-{\mathrm{cos}}^{3}x=\mathrm{cos}\text{\hspace{0.17em}}x\left(1-{\mathrm{cos}}^{2}x\right)$
$=\mathrm{cos}\phantom{\rule{0.2em}{0ex}}x{\mathrm{sin}}^{2}x$

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

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

Answers will vary. Sample proof:
$\frac{1+{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}=\frac{1}{{\mathrm{cos}}^{2}x}+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}={\mathrm{sec}}^{2}x+{\mathrm{tan}}^{2}x={\mathrm{tan}}^{2}x+1+{\mathrm{tan}}^{2}x=1+2{\mathrm{tan}}^{2}x$

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

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

Answers will vary. Sample proof:
${\mathrm{cos}}^{2}x-{\mathrm{tan}}^{2}x=1-{\mathrm{sin}}^{2}x-\left({\mathrm{sec}}^{2}x-1\right)=1-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x+1=2-{\mathrm{sin}}^{2}x-{\mathrm{sec}}^{2}x$

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

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

#### Questions & Answers

find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what