SOLUTION: sin 5pi/12

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Question 623484: sin 5pi/12
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I assume the problem is to find this value exactly. Please include the instructions with the problem when you post.

The only exact Trig values come from the special angles. So to find sin%285pi%2F12%29+we+need+to+find+a+connection+between+%7B%7B%7B5pi%2F12 and one of the special angles: 0, pi%2F6, pi%2F4, pi%2F3, pi%2F2 and any multiple of these.

One connection (maybe not the only one) is that twelfths are half of sixths. So 5pi%2F12 is 1/2 of 5pi%2F6. We can use the sin%28%281%2F2%29x%29 formula:
sin%28%281%2F2%29x%29+=+0+%2B-+sqrt%28%281-cos%28x%29%29%2F2%29
(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.)

Since 5pi%2F12 is less than pi%2F2 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.

Since 5pi%2F12+=+%281%2F2%29%285pi%2F6%29:
sin%285pi%2F12%29+=+sqrt%28%281-cos%285pi%2F6%29%29%2F2%29
Now we just simplify as much as possible. For cos%285pi%2F6%29, 5pi%2F6 terminates in the second quadrant and the reference angle is pi%2F6. cos is negative in the second quadrant and cos%28pi%2F6%29+=+sqrt%283%29%2F2. So cos%285pi%2F6%29+=+-sqrt%283%29%2F2. Inserting this into our formula:
sin%285pi%2F12%29+=+sqrt%28%281-%28-sqrt%283%29%2F2%29%29%2F2%29
sin%285pi%2F12%29+=+sqrt%28%281%2Bsqrt%283%29%2F2%29%2F2%29
sin%285pi%2F12%29+=+sqrt%28%282%2F2%2Bsqrt%283%29%2F2%29%2F2%29
sin%285pi%2F12%29+=+sqrt%28%282%2Bsqrt%283%29%29%2F2%29%2F2%29
sin%285pi%2F12%29+=+sqrt%28%282%2Bsqrt%283%29%29%2F4%29
sin%285pi%2F12%29+=+sqrt%282%2Bsqrt%283%29%29%2Fsqrt%284%29
sin%285pi%2F12%29+=+sqrt%282%2Bsqrt%283%29%29%2F2