Finding points of intersection algebraically between 2 trig functions

• Kitty Kat
In summary: Question 2:sinx = 0 OR cosx - 1 = 0x = 0, pi, 2pi, 3pi, 4piQuestion 3:sinx = 0 OR cosx - 1 = 0x = 0, pi, 2pi, 3pi, 4pi All of them are correct, congratulations!##\sin^2x## or ##(\sin x)^2## are the usual ways of writing ##\sin^2x##, although ##\sin^2x## is far more common.
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

Homework Statement

Find all the points of intersection algebraically of the graphs of ... on the interval [0, 4π]

f(x) = sinx + 1
g(x) = cosx

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

Homework Statement

Find all the points of intersection algebraically of the graphs of ... on the interval [0, 4π]

f(x) = tanx
g(x) = sinx

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

Homework Statement

Find all the points of intersection algebraically of the graphs of ... on the interval [0, 4π]

f(x) = tanx
g(x) = cotx

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.

Any helpful advice or tips would be really appreciated! Thankyou.

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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?

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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).

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##.

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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

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...

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1 person
Kitty Kat said:
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
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 ?

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.

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?

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##.

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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

Kitty Kat said:
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

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.

1. How do I find points of intersection between two trigonometric functions?

To find points of intersection algebraically between two trigonometric functions, you need to set the two functions equal to each other and solve for the variable. The solutions to this equation will be the x-coordinates of the points of intersection. You can then plug these x-values back into either of the original functions to find the corresponding y-values.

2. Can I use the unit circle to find points of intersection between trigonometric functions?

Yes, you can use the unit circle to find points of intersection between trigonometric functions. By plotting both functions on the unit circle, you can visually determine the points of intersection by looking for where the two functions intersect on the circle.

3. Is it possible for two trigonometric functions to have more than one point of intersection?

Yes, it is possible for two trigonometric functions to have more than one point of intersection. This can occur when the two functions have multiple periods or when they have the same period but different phase shifts.

4. What should I do if I can't solve for the variable in the equation?

If you are unable to solve for the variable in the equation, you can use a graphing calculator or a graphing software to graph both functions and find the points of intersection visually. You can also use numerical methods such as Newton's method or the bisection method to approximate the solutions.

5. Can I use the Pythagorean identity to find points of intersection between trigonometric functions?

Yes, you can use the Pythagorean identity (sin^2x + cos^2x = 1) to find points of intersection between trigonometric functions. By manipulating the equations of the two functions to have one of the trigonometric functions in terms of the other, you can then use the Pythagorean identity to solve for the variable.

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