View Full Version : Find exact values for sine, cosine, and tangent of 19pi/12
12-17-2006, 01:37 PM
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.
12-17-2006, 01:59 PM
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 angle-difference 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 angle-difference identities.
. . .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) - (sqrt/2)(1/sqrt)
. . . . .= -1 / (2 sqrt) - sqrt / (2 sqrt)
. . . . .= (-1 - sqrt) / (2 sqrt)
. . . . .= (-sqrt - sqrt) / 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.
12-17-2006, 02:41 PM
Thanks!! That was very helpful.
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