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

3 Answers
Feb 4, 2015

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

Feb 8, 2015

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#

Jun 22, 2016

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)#