Linear approximations derived from Taylor series

[_Greg_]

Homework Statement

So I have the problem questiona dn my teachers solution posted below.
I understand:

f(x_o) = sin pi/6
f '(x_o) = cos pi/6

but i don't know how he gets them into fraction form with the SQRT of 3, it looks like some pythagoras but i don't really know how he did it.

Homework Equations

Question 3(a) (sorry I named it 'answer' instead of 'question')

The Attempt at a Solution

 

[tiny-tim]

Hi _Greg_!

(have a pi: π )

f(x_o) = sin pi/6
f '(x_o) = cos pi/6

but i don't know how he gets them into fraction form with the SQRT of 3, it looks like some pythagoras but i don't really know how he did it.

I can't see the pictures yet, but π/6 = 30º, which is the little angle of half an equilateral triangle, so sin(π/6) (opp/hyp) = 1/2, and yes from Pythagoras you get cos(π/6) = √3/2.

 

[_Greg_]

Thanks for the fast reply tiny tim.
But how would you know its a 30* triangle, and how do you know cos pi/6 = sqrt3/2
Like sin pi/6, is pi close enough to 3 to just call it 3/6?
I don't really know much of the basics, I'm pretty much learning this for the deep end, looking for trends to understand

 

[tiny-tim]

(what happened to that π i gave you? )

But how would you know its a 30* triangle

Because π = 180º, so π/6 = 30º

(and an equilateral triangle has all angles equal, and they add up to 180º, so each angle is 60º, and half that is 30º)

Like sin pi/6, is pi close enough to 3 to just call it 3/6?

No, just draw half an equilateral triangle …

one side will be half the length of the other.

 

[_Greg_]

aaaaaaah, I see, I thought that little n symbol was pi.
And of course, it all adds up to 180*, high school maths is all coming back now
Many thanks