PDA

View Full Version : Find exact values for sine, cosine, and tangent of 19pi/12

smsmith
12-17-2006, 12:37 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.

stapel
12-17-2006, 12:59 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.

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.

smsmith
12-17-2006, 01:41 PM