Question

How do you use the angle sum or difference identity to find the exact value of cos((11pi)/12)#?

Answer 1

`cos((11pi)/12)=-cos(pi/12)`

Explanation:

`cos((11pi)/12)`

= `cos(pi-pi/12)`

Now using identity `cos(A-B)=cosAcosB+sinAsinB`, this becomes

`cospicos(pi/12)+sinpisin(pi/12)`

= `-1xxcos(pi/12)+0xxsin(pi/12)`

= `-cos(pi/12)`

Answer 2

`- sqrt(2 + sqrt3)/2`

Explanation:

Trig unit circle -->
`cos ((11pi)/12) = cos (- pi/12 + pi) = - cos (pi/12)` (1)
Evaluate `cos (pi/12)` by the trig identity:
`2cos^2a = 1 + cos 2a`
`2cos ^2 (pi/12) = 1 + cos (pi/6) = 1 + sqrt3/2 = (2 + sqrt3)/2`
`cos^2 (pi/12) = (2 + sqrt3)/4`
`cos (pi/12) = +- sqrt(2 + sqrt3)/2`
Since `cos (pi/12)` is positive, then only the positive value is accepted.
Finally, from (1):
`cos ((11pi)/12) = - cos (pi/12) = - sqrt(2 + sqrt3)/2`

Check by calculator.
`cos ((11pi)/12) = cos 165 = - 0.97`
`- sqrt(2 + sqrt3)/2 = sqrt(3.73)/2 = 1.93/2 = 0.97.` OK