Question

How do you find the exact values of the sine, cosine, and tangent of the angle `(-7pi)/12`?

Answer 1

Call `t = (-7pi)/12`
On the unit circle, we have:
`sin t = - cos (pi/12)` (1)
`cos t = - sin (pi/12)` (2)
Next, find `sin (pi/12) and cos (pi/12)` by using trig identities:
sin (a - b) = sin a.cos b - sin b.cos a
cos (a -b) = cos a.cos b + sin a.sin b
In this case
#sin (pi/12) = sin (pi/3 - pi/4) = (sqrt3/2)(sqrt2/2) - (sqrt2/2)(1/2) = (sqrt6 - sqrt2)/4 #
`cos (pi/12) = cos (pi/3 - pi/4) = (sqrt6 + sqrt2)/4`
From (1) and (2), we get:
`sin t = - cos (pi/12) = - (sqrt6 + sqrt2)/4`
`cos t = - sin (pi/12) = - (sqrt6 - sqrt2)/4`
From there,
`tan t = (sqrt6 + sqrt2)/(sqrt6 - sqrt2)`
`cot t = (sqrt6 - sqrt2)/(sqrt6 + sqrt2)`
`sec t = 1/(sin t) = - 4/(sqrt6 + sqrt2)`
`csc t = 1/(cos t) = - 4/(sqrt6 - sqrt2)`