B Cosine of 1 degree and cosine of 60 degrees?

[symbolipoint]

Yes, you are obsessed with plotting graphs, all your questions are about plotting graphs but you cannot learn anything this way.

If you don't understand why the base of that triangle is $\sqrt{3}$ long then you are not ready for sine and cosine yet, you need to start with Pythagoras. This course should do:
https://www.khanacademy.org/math/geometry/hs-geo-trig

So this points back to some of the simpler Geometry you study in high school, about right triangles, some regular polygons, equilateral triangles.

 

[symbolipoint]

Take an equilateral triangle of side 2. Cut it in half along a bisector.

I think you really meant "sides o three"? Two of the sides become hypotenuses of the two new triangles after bisecting the equilat. into the two right triangles.

 

[Mark44]

Take an equilateral triangle of side 2. Cut it in half along a bisector.

I think you really meant "sides o three"? Two of the sides become hypotenuses of the two new triangles after bisecting the equilat. into the two right triangles.

No, @pasmith meant what he wrote; namely, an equilateral triangle whose sides are of length 2.

The angle at the lower left is 60°, so $\cos(60°) = \frac 1 2$.

 

[anuttarasammyak]

In a right-angled triangle 30 - 60 - 90, why adjacent side is given 1√3 unit, its hypotenuse 2 units and its opposite side is 1 unit?

Pytagoras's theorem
a^2+b^2=c^2
says for this case
1^2+\sqrt{3}^2=2^2
You can easily find proof of Pytagoras's theorem on Web.

You can make a equilateral triangle from the two of these triangles then you will find the angles are 30- 60- 90 as shown in post #33.

 

[anuttarasammyak]

Why is cos (1)° = 0.9998?

If you have got understood the formula
\cos2\theta=2cos^2\theta-1
we can make use of it sequentially to get half-angle, quarter angle, one eighth angle,... of cosine.
The proof of the formula is easily found in the web.

Starting from cos 60 degree = 1/2, cos30, cos15, cos(15/2), cos(15/4),cos(15/8), cos(15/16)

Thus we can calculate value of cos (15/16) degree and may hope it an approximate value of cos 1 degree. Please try it if you have an interest in this method.