# Homework Help: Finding points of intersection algebraically between 2 trig functions

1. Jul 15, 2014

### Kitty Kat

So I have several problems that ask me to find all points of intersection algebraically, but I haven't been able to make much headway on most of them.

The first problem
1. The problem statement, all variables and given/known data
Find all the points of intersection algebraically of the graphs of .... on the interval [0, 4π]

2. Relevant equations
f(x) = sinx + 1
g(x) = cosx

3. The attempt at a solution

-Attempt #1:
sinx + 1 = cosx
sinx + 1 - cosx = 0
sinx - cosx = -1

-Attempt #2:
sinx + 1 = cosx
sinx / cosx + 1 / cosx = cosx / cosx
tanx + secx = 1
sinx/cosx + 1/cosx = 1

-Attempt #3:
sinx + 1 = cosx
sinx^2 + 1 = cosx^2
sinx^2 + 1 = 1 - sinx^2
2sinx^2 + 1 = 1
2sinx^2 = 0

... I think I'm really close to the answer here, but I'm not sure where to go now.

The second problem
1. The problem statement, all variables and given/known data
Find all the points of intersection algebraically of the graphs of .... on the interval [0, 4π]

2. Relevant equations
f(x) = tanx
g(x) = sinx

3. The attempt at a solution

sinx = tanx
sinx = sinx / cosx
sinx * cosx - sinx = 0
sinx (cosx - 1) = 0

sinx = 0
cosx = 1

y = 0
x = 1
(1,0)

It's (1,0) at [0, 2π, 4π], but apparently [π, 3π] are also points of intersection, even though the cosine value at [π, 3π] are -1. Could somebody explain that please? (I've pretty much already solved it)

The third problem
1. The problem statement, all variables and given/known data
Find all the points of intersection algebraically of the graphs of .... on the interval [0, 4π]

2. Relevant equations
f(x) = tanx
g(x) = cotx

3. The attempt at a solution
Attempt #1:
tanx = cotx
sinx / cosx = cosx / sinx
sinx^2 / cosx = cosx
sinx^2 = cosx^2
sinx^2 = -sinx^2 + 1
2sinx^2 = 1

And again ... I think I'm really close to the answer here, but I'm not sure where to go now.

Last edited: Jul 15, 2014
2. Jul 15, 2014

### HakimPhilo

Question 1:
You used the correct method to obtain the result (which is to square both sides and then substitute $\cos^2x$ for $1-\sin^2x$), however you made a mistake when simplifying since $(\sin x+1)^2\neq \sin^2 x+1$. To obtain the correct result you may use the following identity: $(a+b)^2=a^2+2ab+b^2$.

Question 2:
$\sin x=0\,\text{ or }\,\cos x=1$ does not mean that $y = 0, x = 1$, can you see where you made the mistake?

Question 3:
Let me help you again: $2\sin^2x = 1\Rightarrow 2\sin^2x-1=0\Rightarrow\left(\sqrt{2}\sin x-1\right)\left(\sqrt{2}\sin x+1\right)=0.$ Do you know how to proceed?

Last edited: Jul 15, 2014
3. Jul 15, 2014

### Kitty Kat

Thanks a lot HakimPhilo, you really cleared it up for me, but I still don't understand question 2 though.
I thought that the only time cosx = 1 was when x = 1 (Likewise for sinx = 0 being y = 1).

4. Jul 15, 2014

### HakimPhilo

You're welcome. For Q.2 you should understand that we're dealing with units of measure of angles. So when you say $\cos x=1$ means $x=1$ you should specify the unit you used, and since you're surely dealing with radians then for $x=1$ doesn't mean that $\cos x=1$. (try it on your calculator)
Instead of $x=1$, calculate the trigonometric values of an angle with radian measure $0$.

Last edited: Jul 15, 2014
5. Jul 15, 2014

### Kitty Kat

Mmm, I think I'm still a little slow on the uptake. I still don't really understand why [π, 3π] work, because the cos values of [π, 3π] are both -1, and if cosx = 1, then that doesn't make the equation true, since -1 doesn't equal 1. I think I'm really missing some fundamental basic idea here though ....

cosx = 1
cos(pi) = -1
cos(3pi) = -1

6. Jul 15, 2014

### HakimPhilo

Do you know how to solve equations of the form $a\cos x+b=0$ or $a\sin x+b=0$? If not then the following may help you a bit:
$a\cos x=b\Rightarrow \cos x=\tfrac ba.$ What this equation equation means is that we're looking for a number $x$ such that $\cos x$ is equal to $b/a$. To find this number we can use our classical trigonometric values table:

http://hotmath.com/hotmath_help/topics/values-of-trigonometric-ratios-for-standard-angles/values-of-trigonometric-ratios-for-standard-angles-image004.gif

So for example if we want to solve the equation $2\cos x=\sqrt{2}$, we proceed as follows:
$2\cos x=\sqrt{2}\Rightarrow \cos x=\tfrac{\sqrt{2}}2.$ We look at our trigonometric table and we find that the angle whose cosine corresponds to $\tfrac{\sqrt{2}}2$ is $\tfrac\pi4$. So $\tfrac\pi4$ solves our equation. But it's not the only solution, since numbers of the form $\tfrac\pi4+2n\pi$ (where $n$ is an integer) are also solutions.

If it is not in our table then we use the $\arccos$ function. For example if you have an equation like $3\cos x=1\Rightarrow \cos x=\tfrac13\,(1)$, you can see that $\tfrac13$ isn't in our table. To solve this problem we use the $\arccos$ function which is like the inverse of $\cos$, more precisely $\arccos(\cos x)=x$. So by $(1)$ we have that $\arccos(\cos x)=\arccos \tfrac13\Rightarrow x=\arccos \tfrac13$. And remember not only $\arccos \tfrac13$ is a solution but also numbers of the form $\arccos \tfrac13+2n\pi$. (where $n$ is an integer)

If you need more explanations feel free to ask.

For the moment I have to sleep...

Last edited: Jul 15, 2014
7. Jul 15, 2014

### SammyS

Staff Emeritus
This is for your problem #2, right?

You have an equation which you show is equivalent to

sin(x) (cos(x) - 1) = 0 .

This equation is satisfied if either of the following conditions is is true.

sin(x) = 0 OR cos(x) - 1 = 0.

It doesn't matter that cos(π) = -1 , because sin(π) = 0 .

What is 0 times -2 ?

8. Jul 15, 2014

### Kitty Kat

Thank you Sammy for clearing that up! I thought it had to satisfy both equations, since I was finding the point of intersection, but I guess it only needs to satisfy one or the other.

And thanks again HakImPhilo for the in depth explanation, it really helped clear my understanding.

9. Jul 16, 2014

### Kitty Kat

Mmm, so just to double check my answers...

Question 1:
sinx + 1 = cosx
(sinx + 1)^2 = 1 - sin^2x
sin^2x + 2sinx + 1 = 1 - sin^2x
2sin^2x + 2sinx = 0
sin^2x + sinx = 0

x = 0, pi, 2pi, 3pi, 4pi

(And for the record, how would I input sin^2x into my calculator anyways? I think it's different from sinx^2 right?)

Question 3:

$2\sin^2x = 1\Rightarrow 2\sin^2x-1=0\Rightarrow\left(\sqrt{2}\sin x-1\right)\left(\sqrt{2}\sin x+1\right)=0.$

sinx = -1 / sqrt(2) → sinx = -sqrt(2) / 2
sinx = +1 / sqrt(2) → sinx = sqrt(2) / 2

x = 5pi/4, 7pi/4, 13pi/4, 15pi/4, 3pi/4, pi/4, 11pi/4, 9pi/4

--------------------------------------------------------
And on an unrelated note, how would I graph something like f(x) = cos(x+2)? I understand how to graph f(x) = cos(x), but the +2 is really throwing me off (I also understand that it shifts the graph horizontally right by 2, but it feels really awkward graphing it, considering my x-axis goes from 0 - 2pi, in increments of pi/2). Should I just try to approximate it?

10. Jul 16, 2014

### HakimPhilo

Question 1:
Try plugging in $0,\pi,2\pi,3\pi,4\pi$ and see if they satisfy your original equation $\sin x+1=\cos x$.

You should distinguish between $\sin^2x$ and $\sin x^2$. The first one is just a notation for $(\sin x)(\sin x)$, so if you want to calculate it you first find the value of $\sin x$ then you square it. The second one means you calculate the sine of x squared.

Question 2:
Yes, all the numbers you listed are solutions to that equation that are in the interval $[0, 4π]$.

_________________________________________________________

Do you know how to graph a function of the form $f(x)=A\cos(Bx-C)$ in general? If not then the following may help you:

To see the difference between $f(x)=\cos x$ and $g(x)=\cos (x+L)$, I've ploted this graphic for various values of $L$. And as you can see, they shift the graph of $\cos x$ by a horizontal amount of $L$.

Last edited: Jul 16, 2014
11. Jul 16, 2014

### Kitty Kat

Wow, thankyou again HakImPhilo, that diagram really helped!

Am I still doing problem 1 incorrectly? I seem to get different points of intersection when I graph
sin^2x + sinx = 0
compared to sinx + 1 = cosx

sinx + 1 = cosx
(sinx + 1)^2 = 1 - sin^2x
sin^2x + 2sinx + 1 = 1 - sin^2x
2sin^2x + 2sinx + 1 = 1
2sin^2x + 2sinx = 0
sin^2x + sinx = 0

12. Jul 16, 2014

### HakimPhilo

That's because when you squared both sides of the equations you added extraneous solutions. So you'll have to check the solutions you got and to eliminate those that don't work.