How do you find `sin(pi/12)` and `cos(pi/12)`?

I would use the expansion in series of the two functions, as

(have a look at the page: http://en.wikipedia.org/wiki/Taylor_series for more info)

Where a function (in a point) is given by an infinite sum of values.
The `n!` is called "factorial" and `x` is in radians.

We choose few values only, depending upon the accuracy we want (basically, decimal digits you want).
For your case (3 decimals only):

`sin(pi/12)=pi/12-(pi/12)^3/(3*2*1)+(pi/12)^5/(5*4*3*2*1)-....`
`=pi/12-1/6(pi)^3/(12^3)+1/120(pi^5)/(12^5)-...=0.261-0.003+0.000...=`

`=0.258`

Now you can try to do the same by yourself with `cos` (which starts at `1`).

hope it helps

Here is another way to solve this problem.

It's known that `sin^2(phi)+cos^2(phi)=1` for any angle `phi`.
Therefore,
`sin^2(pi/12)+cos^2(pi/12)=1`

Let's use a formula for a sine of a double angle:
`sin(2phi)=2*sin(phi)*cos(phi)`

Using this formula,
`2*sin(pi/12)*cos(pi/12)=sin(2*pi/12)=sin(pi/6)=1/2`

Substitute for simplicity:
`x = cos(pi/12)` and `y = sin(pi/12)`
Both are positive (since an angle `pi/12` is in the first quadrant).
Also `x>y` since, as angle `phi` increases from 0 to `pi/12`, `cos(phi)` decreases from `1` and `sin(phi`) increases from `0`. They meet only at `pi/4`, so at `pi/12` function `cos` is greater than function `sin`.

We have a system of two equations with two unknowns:
`x^2+y^2=1`
`2xy=1/2`

Adding the second equation to the first, we get
`x^2+2xy+y^2=3/2` or
`(x+y)^2=3/2`
Since both `x` and `y` are positive
`x+y=sqrt(3/2)=sqrt(6)/2`

Subtracting the second equation from the first, we get
`x^2-2xy+y^2=1/2` or
`(x-y)^2=1/2`
Since `x>y`
`x-y=sqrt(1/2)=sqrt(2)/2`

So, we have a very simple system of two equations with two unknowns:
`x+y=sqrt(3/2)=sqrt(6)/2`
`x-y=sqrt(1/2)=sqrt(2)/2`

Adding and subtracting this equations, we find solutions:
`2x=(sqrt(6)+sqrt(2))/2`
`2y=(sqrt(6)-sqrt(2))/2`

Solutions are
`cos(pi/12)=x=(sqrt(6)+sqrt(2))/4`
`sin(pi/12)=y=(sqrt(6)-sqrt(2))/4`

Use the cosine and sine half angle formulas:

• `sin(theta/2)=+-sqrt((1-cos(theta))/2)`
• `cos(theta/2)=+-sqrt((1+cos(theta))/2)`

First, let's solve for `sin(pi/12)`. If we let `theta=pi/6`, then we see that:

`sin((pi/6)/2)=+-sqrt((1-cos(pi/6))/2)`

We will take the positive root since the angle is `(pi/6)/2=pi/12`, which is in the first quadrant, where sine is positive.

`color(blue)(sin(pi/12))=sqrt((1-sqrt3/2)/2)=sqrt((2-sqrt3)/4)color(blue)(=sqrt(2-sqrt3)/2`

The cosine method is almost identical. The positive root will again be taken because cosine is also positive in the first quadrant.

`cos((pi/6)/2)=sqrt((1+cos(pi/6))/2)`

Hence:

`color(red)(cos(pi/12))=sqrt((1+sqrt3/2)/2)=sqrt((2+sqrt3)/4)color(red)(=sqrt(2+sqrt3)/2)`