to the curve at the origin make with the x-axis?
I found that so when ,
I feel like I should be ending up with
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.
I am following along but for whatever reason it is still not clicking. I understand 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 to the theta
I know once I get it it'll be a duh moment.
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
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?