Finding the intersection points on the graph y=sinx, y=cosx and y=tanx

[Alexander1]

Hi guys, I'm new to this site and it seems like it will be a great resource when I'm stuck on a problem. I'll firstly set out the question and then add in my working so far.

Question: I was firstly asked to graph the trigonometric functions y=sinx, y=cosx and y=tanx in the interval where x is greater than or equal to 0 but less than or equal to pi/2. My question is to find algebraically the points of intersection of the three curves in that interval. So far I have found the intersection points for where y=cosx and y=sinx cross (hopefully I'm right - I got root 2/2 for both coordinates).

I'm now struggling with finding the intersection points for where y=cosx and y=tanx cross as well as the points for where y=sinx and y=tanx cross.

I know that the points for y=sinx and y=tanx will be (0,0) as it can easily be see from the graph. This is my working so far for that...

1. y=sinx
2. y=tanx

Sub 1. into 2.
sinx=tanx

Using the identity:
tanx=sinx/cosx

(I'm not too sure here if I'm meant to be finding an angle?)
sinθ=sinθ/cosθ
cosθ x sinθ = sinθ
cosθ=1
θ=cos-1 (1)
θ=0

I'm not really sure where to go from here to show that it must have an intersection point of (0,0) - It would be greatly appreciated if someone could please help me on this.

Lastly, I'm also not sure how to find the intersection points for where y=cosx and y=tanx cross.

For this I really haven't got much at all. I do know that I must use the identity sin2 x + cos2 x = 1 (I'm pretty sure). At the moment this is all I have...

1. y=cosx
2. y=tanx

sub 1. into 2.
cosx=tanx
cosx=sinx/cosx
cosx x cosx = sinx
(This step below may be wrong?)
cos2x=sinx

Using the identity sin2 x + cos2 x = 1
cos2 x=1-sin2 x

From here I'm stuck on how to find the intersection points?

If anyone could help me find these intersection points for the 2 questions above I would be very grateful. Sorry about the lengthy thread.

Thanks in advance!

 

[anemone]

Hi guys, I'm new to this site and it seems like it will be a great resource when I'm stuck on a problem. I'll firstly set out the question and then add in my working so far.

Question: I was firstly asked to graph the trigonometric functions y=sinx, y=cosx and y=tanx in the interval where x is greater than or equal to 0 but less than or equal to pi/2. My question is to find algebraically the points of intersection of the three curves in that interval. So far I have found the intersection points for where y=cosx and y=sinx cross (hopefully I'm right - I got root 2/2 for both coordinates).

Hello bersim! Welcome to MHB!

Yes, you're right, $\sin x=\cos x$ yields the intersection point $\left(\dfrac{\pi}{4},\,\dfrac{\sqrt{2}}{2}\right)$.

I'm now struggling with finding the intersection points for where y=cosx and y=tanx cross as well as the points for where y=sinx and y=tanx cross.

I know that the points for y=sinx and y=tanx will be (0,0) as it can easily be see from the graph. This is my working so far for that...

1. y=sinx
2. y=tanx

Sub 1. into 2.
sinx=tanx

Using the identity:
tanx=sinx/cosx

(I'm not too sure here if I'm meant to be finding an angle?)

Yes, in solving for any trigonometric functions, we are actually dealing with the angles and so, our $x$ is angle measured in radians.

sinθ=sinθ/cosθ
cosθ x sinθ = sinθ
cosθ=1
θ=cos-1 (1)
θ=0

I'm not really sure where to go from here to show that it must have an intersection point of (0,0) - It would be greatly appreciated if someone could please help me on this.

Here, I see that you divided the equation $\cos x \sin x= \sin x$ by $\sin x$. You could do that if $\sin x\ne 0$ because we can never divide any number by zero, as division by zero is undefined.

So, what we could do is, we move $\sin x$ to the left side, and then we factor the LHS expression.

$\cos x \sin x= \sin x$

$\cos x \sin x- \sin x=0$

$\sin x(\cos x-1) =0$

When we have a product equals zero, then either one of them or two of them is zero.

We hence set $\sin x=0\implies x=0$ on the given interval and $\cos x-1=0\implies x=0$ and so, when $x=0$, $y=\sin x=\sin 0=0$ and the only intersection point between the graph $y=\sin x$ and $y=\tan x$ on the given interval is $(0,\,0)$.

Lastly, I'm also not sure how to find the intersection points for where y=cosx and y=tanx cross.

For this I really haven't got much at all. I do know that I must use the identity sin2 x + cos2 x = 1 (I'm pretty sure). At the moment this is all I have...

1. y=cosx
2. y=tanx

sub 1. into 2.
cosx=tanx
cosx=sinx/cosx
cosx x cosx = sinx
(This step below may be wrong?)
cos2x=sinx

Using the identity sin2 x + cos2 x = 1
cos2 x=1-sin2 x

From here I'm stuck on how to find the intersection points?

What you need to do now is to replace $\cos^2 x=1-\sin^2 x$ into the equation $\cos^2 x=\sin x$ and solve for $\sin x$.

Can you show me how you would do in this last part of the problem?

 

[Alexander1]

Hi Anemone, thanks very much for your help. I actually managed to find all of the intersection points after a bit of thinking. For the y=cosx and y=tanx one I used the quadratic formula and then found the 2 x intercepts, although only one is valid due to it being in the 1st quadrant. I then subbed x into y=cosx (although it works with both) and found my y coordinate.

Cheers, Jack

 

[anemone]

I am happy that you solved the problem! I welcome you to ask(post) again when there is something in math that puzzles you.

 

[Alexander1]

I'll definitely make sure I use this again when I'm having trouble! It seems so obvious now haha.

Thanks