Question 623484
I assume the problem is to find this value exactly. Please include the instructions with the problem when you post.<br>
The only exact Trig values come from the special angles. So to find {{{sin(5pi/12) we need to find a connection between {{{5pi/12}}} and one of the special angles: 0, {{{pi/6}}}, {{{pi/4}}}, {{{pi/3}}}, {{{pi/2}}} and any multiple of these.<br>
One connection (maybe not the only one) is that twelfths are half of sixths. So {{{5pi/12}}} is 1/2 of {{{5pi/6}}}. We can use the {{{sin((1/2)x)}}} formula:
{{{sin((1/2)x) = 0 +- sqrt((1-cos(x))/2)}}}
(Note: The formula does not usually have the zero. I had to put it there because algebra.com's formula drawing software requires a number in front of a "plus or minus" sign.)<br>
Since {{{5pi/12}}} is less than {{{pi/2}}} it terminates in the first quadrant. Since sin is positive in the first quadrant we know to use the positive square root. So I am going to dispense with the "plus or minus" sign.<br>
Since {{{5pi/12 = (1/2)(5pi/6)}}}:
{{{sin(5pi/12) = sqrt((1-cos(5pi/6))/2)}}}
Now we just simplify as much as possible. For {{{cos(5pi/6)}}}, {{{5pi/6}}} terminates in the second quadrant and the reference angle is {{{pi/6}}}. cos is negative in the second quadrant and {{{cos(pi/6) = sqrt(3)/2}}}. So {{{cos(5pi/6) = -sqrt(3)/2}}}. Inserting this into our formula:
{{{sin(5pi/12) = sqrt((1-(-sqrt(3)/2))/2)}}}
{{{sin(5pi/12) = sqrt((1+sqrt(3)/2)/2)}}}
{{{sin(5pi/12) = sqrt((2/2+sqrt(3)/2)/2)}}}
{{{sin(5pi/12) = sqrt((2+sqrt(3))/2)/2)}}}
{{{sin(5pi/12) = sqrt((2+sqrt(3))/4)}}}
{{{sin(5pi/12) = sqrt(2+sqrt(3))/sqrt(4)}}}
{{{sin(5pi/12) = sqrt(2+sqrt(3))/2}}}