Question

How do you find the exact functional value sec (19Pi/12) using the cosine sum or difference identity?

Answer 1

Find `sec ((19pi)/12)`

Ans: `- sqrt(2 + sqrt3)/2`

Explanation:

Call `(19pi/12) = t` --> `sec t = 1/sin t.`
`cos 2t = cos ((19pi)/6) = cos ((7pi)/6 + 2pi) = -cos pi/6 = -sqrt3/2`
Use the trig identity: `cos 2t = 1 - 2sin^2 t.`
cos 2t =`-sqrt3/2 = 1 - 2sin^2 t.`
2sin^2 t = 1 + sqrt3/2 = (2 + sqrt3)/2
sin^2 t = (2 + sqrt3)/4
sin t = `sin ((19pi)/12) = +- sqrt(2 + sqrt3)/2`
Since the arc `((19pi)/6)`= 285 deg is located in Quadrant IV, its sin is negative.
Then, `sin ((19pi)/12) = - sqrt(2 + sqrt3)/2`

Check by calculator: x = 285 --> sin 285 = -0.97.
`- sqrt(2 + sqrt3)/2 = -1.931/2 = -0.97` OK