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

##### 3 Answers

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

We choose few values only, depending upon the accuracy we want (basically, decimal digits you want).

For your case (3 decimals only):

Now you can try to do the same by yourself with

hope it helps

Here is another way to solve this problem.

It's known that

Therefore,

Let's use a formula for a sine of a double angle:

Using this formula,

Substitute for simplicity:

Both are positive (since an angle

Also

We have a system of two equations with two unknowns:

Adding the second equation to the first, we get

Since both

Subtracting the second equation from the first, we get

Since

So, we have a very simple system of two equations with two unknowns:

Adding and subtracting this equations, we find solutions:

Solutions are

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

We will take the positive root since the angle is

#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)#