# Thread: what angle does the tangent line

1. ## what angle does the tangent line

to the curve $\displaystyle y = \frac{1}{\sqrt3} \sin 3x$at the origin make with the x-axis?

I found that $\displaystyle y' = \sqrt3 \cos 3x$ so when $\displaystyle x = 0$, $\displaystyle \sqrt3 \cos 3(0) = \sqrt3*1 = \sqrt3$

I feel like I should be ending up with $\displaystyle \frac{\sqrt3}{2}$

2. ## Re: what angle does the tangent line

Originally Posted by Jonroberts74
to the curve $\displaystyle y = \frac{1}{\sqrt3} \sin 3x$at the origin make with the x-axis?

I found that $\displaystyle y' = \sqrt3 \cos 3x$ so when $\displaystyle x = 0$, $\displaystyle \sqrt3 \cos 3(0) = \sqrt3*1 = \sqrt3$

I feel like I should be ending up with $\displaystyle \frac{\sqrt3}{2}$
The x axis is when y=0
when y=0 x=0, pi/3, 2pi/3 ..... npi/3 where n is a positive integer.

So x =0 will only give 1 of the answers. And I got root3, the same as you.
When tan theta = root3 theta=pi/3 radians (angle on the right between the x axis and the tangent)

You will now need to find y' for x=pi/3. (you should get -root3 I think)
The angle will be the same but it will be on the left side. The angle between the positive side will be 2pi/3

And, you will need to convince yourself that the others are a repeat of these 2.

If you think I've done something wrong or you would like more of an explanation, just ask.
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When you are working with graphs it always makes it easier if you have the best idea possible of what the graph looks like. (Without doing too much work)
This is a sine graph (oscilating about the x axis)
with amplitude 1/root3
and period = 2pi/3
A knowledge of the graph will tell you that it will cross the x axis at 0,pi/3, 2pi/3 etc
and it will have a positive gradient at x=0 a negative one at x=pi/3 and these 2 gradients will repeat themselves with each intersection of the x axis.

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I appologize for the presentation of this answer, I don't know haw to use latex yet. I've just started learning but I am having difficulty with it.

3. ## Re: what angle does the tangent line

Originally Posted by Jonroberts74
to the curve $\displaystyle y = \frac{1}{\sqrt3} \sin 3x$at the origin make with the x-axis?
I found that $\displaystyle y' = \sqrt3 \cos 3x$ so when $\displaystyle x = 0$, $\displaystyle \sqrt3 \cos 3(0) = \sqrt3*1 = \sqrt3$
I feel like I should be ending up with $\displaystyle \frac{\sqrt3}{2}$
What if I told you that $\displaystyle \theta =\frac{\pi}{3}~?$

Recall that $\displaystyle y'$ is the slope and slope is the tangent of the angle the line makes with the positive x-axis.

4. ## Re: what angle does the tangent line

I am following along but for whatever reason it is still not clicking. I understand $\displaystyle \sqrt3$ is the slope because the first derivative is the tangent line, and makes an angle with the x-axis.

I am reviewing my trigonometric functions and identities.

But for some reason I am missing on how to get from the tangent $\displaystyle \sqrt3$ to the theta $\displaystyle \pi/3$

I know once I get it it'll be a duh moment.

5. ## Re: what angle does the tangent line

Originally Posted by Jonroberts74
I am following along but for whatever reason it is still not clicking. I understand $\displaystyle \sqrt3$ is the slope because the first derivative is the tangent line, and makes an angle with the x-axis.
But for some reason I am missing on how to get from the tangent $\displaystyle \sqrt3$ to the theta $\displaystyle \pi/3$.
What is $\displaystyle \tan \left( {\frac{\pi }{3}} \right)=~?$

6. ## Re: what angle does the tangent line

$\displaystyle tan \frac{\pi}{3} =\sqrt3$

Thank you,

I don't feel convinced of this answer, I understand it is fact, but I feel like that is just me memorizing it.

7. ## Re: what angle does the tangent line

It is not a matter of "memorizing" anything to know that if y= mx+ b then for 2 points $\displaystyle (x_0, mx_0+ b)$ and $\displaystyle (x_1, mx_1+ b)$ then $\displaystyle \dfrac{(mx_1+ b)- (mx_0+ b){x_1- x_0}= m$. You might consider it "memorizing" to know that [tex]tan(\theta)[tex] equals "opposite side over near side" so that $\displaystyle tan(\theta)= m$.

8. ## Re: what angle does the tangent line

I understand that,

I mean, Deriving root3 equals tan pi/3 rather than just memorizing it.

9. ## Re: what angle does the tangent line

Originally Posted by Jonroberts74
I understand that,
I mean, Deriving root3 equals tan pi/3 rather than just memorizing it.
It is not about memorizing much. Here is advice at least two generations of undergraduates in complex variables:
Any triangle with sides $\displaystyle 1,~2,~\&~\sqrt3$ is a $\displaystyle \tfrac{\pi}{6},~\tfrac{\pi}{3},~\tfrac{\pi}{2}$ triangle, (30.60,90)
From that you can figure out the trig values for those numbers.

10. ## Re: what angle does the tangent line

You should remember that 30 degrees , 60 degrees and 45 degrees are 'special' angles where you can get exact ratios but other than that I do not remember what they are. I derive them every time but I do it so quickly that it doesn't slow me down.

Draw an eqilateraral triangle and let every side be 2 units long.
Now drop a perpedicular from the apex angle down to the base.
This will cut the apex angle in half giving you 2 congruent triangles with angles 60 degrees, 30 degrees and 90degrees ( 2pi/3, pi/3 and pi/2 radians)
The hypotenuse of each triangle is 2
The base of each is 1
using pythagoras you get the third as $\displaystyle \sqrt{3}$

Now you can read sine, cos and tan of 2pi/3 and pi/3 straight of the diagram.

Is this what you needed to know?