Math (Trigonometry)
posted by Tom .
What is the maximum value of
[9sinθ+2sin(θ+π/3)]^2?

let y = [9sinθ+2sin(θ+π/3)]^2
dy/dØ = (9sinθ+2sin(θ+π/3)) (9cosØ + 2cos(Ø+π/3)
= 0 for a max/min
so (9sinθ+2sin(θ+π/3)) =0 OR (9cosØ + 2cos(Ø+π/3)) = 0
(Where are you getting these questions from, they are tediously long....)
Case1:
9sinØ + 2sin(Ø+π/3) = 0
9sinØ + 2(sinØcosπ/3 + cosØsinπ/3) = 0
9sinØ + 2( (1/2)sinØ + (√3/2)cosØ ) = 0
9sinØ + sinØ + √3cosØ = 0
10sinØ = √3cosØ
sinØ/cosØ = √3/2
tanØ = √3/2
Ø could be in II or IV
in II sinØ = √3/√103 and cosØ = 10/√103
but remember I showed that
9sinØ + 2sin(Ø+π/3)
= 10sinØ + √3cosØ
then y = [9sinØ + 2sin(Ø+π/3)]^2
= (10sinØ + √3cosØ)^2
= (10√3/√103 + √3(10/√103))^2
= 0 , how about that ?
in IV sinØ = √3/√103 and cosØ = 10/√103
and (10sinØ + √3cosØ)^2
= 0 as well
HALF WAY DONE
Case 2:
9cosØ + 2cos(Ø+π/3) = 0
9cosØ + 2(cosØcos π/3  sinØsin π/3) = 0
9cosØ + 2( (1/2)cosØ  (√3/2)sinØ ) = 0
9cosØ + cosØ  √3sinØ = 0
10cosØ = √3sinØ
10/√3 = sinØ/cosØ
tanØ = 10/√3
Ø could be in I or III
in I , sinØ = 10/√103 , cosØ = √3/√103
again, recall that
9sinθ+2sin(θ+π/3) = 10sinØ +√3cosØ
y = (10(10/√103) + √3(√3/√103)^2
= (100/√103 + 3/√103)^2
= (103/√103)^2
= 10000/103 = appr 97.09
in III , sinØ = √3/√103 , cosØ = 10/√103
y = same as above, except
= (103/√103)^2 = 10000/103 = appr 97.09
so the max is 97.09 , and the minimum is 0
Wolfram appears to confirm my answer
http://www.wolframalpha.com/input/?i=plot+%5B9sinθ%2B2sin%28θ%2Bπ%2F3%29%5D%5E2 
Thank you very much! You're good! :)