
New Member
Find exact values for sine, cosine, and tangent of 19pi/12
Help! I am having the toughest time figuring this one out.
Find the exact values for the sine, the cosine, and the tangent of (19pi) / 12
I know that it is equal to (11pi/6)  (pi/4). But what do I do with (11pi/6)? I know the reference angle for (11pi/6) is (pi/6) but what do I do with that? Any help would be appreciated.

Elite Member
Since trig functions are periodic, you can and and subtract multiples of 2pi to, in effect, move the given angle measure back into the "first" period. In your case, (19pi)/12 is already in the first period, since 19/12 < 24/12 = 2.
Since (19pi)/12 = (11pi)/6  pi/4, then you can replace (19pi)/12 with (11pi)/6 pi/4, and apply the angledifference identities.
Since the reference angle for (11pi)/6 is pi/6, you can replace (11pi)/6 with pi/6, taking care with the signs, when you evaluate the expressions derived from the angledifference identities.
For instance:
. . .sin((19pi)/12) = sin((11pi)/6  pi/4)
. . . . .= sin(11pi/6) cos(pi/4)  cos(11pi/6) sin(pi/4)
. . . . .= [sin(pi/6)] cos(pi/4)  cos(pi/6) sin(pi/4)
. . . . .= sin(pi/6) cos(pi/4)  cos(pi/6) sin(pi/4)
. . . . .= (1/2)(1/sqrt[2])  (sqrt[3]/2)(1/sqrt[2])
. . . . .= 1 / (2 sqrt[2])  sqrt[3] / (2 sqrt[2])
. . . . .= (1  sqrt[3]) / (2 sqrt[2])
. . . . .= (sqrt[2]  sqrt[6]) / 4
There are probably other ways to proceed....
Note: You can check your answers in your calculator, by evaluating sin(19pi/12) directly, and comparing the value with the final expression above.
Eliz.

New Member
Thanks!! That was very helpful.
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