Question

Find the exact value of sin(pi/12) , cos (11pi/12) , and tan (7pi/12)?

Answer 1

Find sin (pi/12); cos (pi/12) and tan ((7pi)/12)

Explanation:

Call `sin pi/12 = sin t.` Use trig identity: `cos 2a = 1 - sin^2 a`
`cos 2pi/12 = cos pi/6 = sqrt3/2 = 1 - sin^2 t.`
`sin^2 t = 1 - sqrt3/2 = (2 - sqrt3)/4`
`sin t = +- sqrt(2 - sqrt3)/2`. Since `pi/12` has its sin positive, then
`sin (pi/12) = sin t = sqrt(2 - sqrt3)/2`

Call `cos (pi/2) = cos t`
`cos ((2pi)/12) = sqrt3/2 = 2cos^2 t - 1`
`cos^2 t = (2 + sqrt3)/4`
`cos t = +- sqrt(2 + sqrt3)/2`. Since cos `(pi/12)` is positive, then
`cos (pi/12) = cos t = sqrt(2 + sqrt3)/2`

`tan ((7pi)/12) = tan (pi/12 + pi) = tan (pi/12) = sin/(cos) =`
`= (sqrt(2 - sqrt3))/(sqrt(2 + sqrt3))`