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July 24, 2014

July 24, 2014

Posted by **Tom** on Thursday, May 9, 2013 at 9:15pm.

[9sinθ+2sin(θ+π/3)]^2?

- Math (Trigonometry) -
**Reiny**, Thursday, May 9, 2013 at 11:12pmlet 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

- Math (Trigonometry) -
**Tom**, Friday, May 10, 2013 at 12:06amThank you very much! You're good! :)

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