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! :)
To find the maximum value of the given expression, we can use the concept of trigonometric identities.
Step 1: Expand the given expression.
[9sinθ+2sin(θ+π/3)]^2 = [9sinθ+2sinθcos(π/3)+2cosθsin(π/3)]^2
Step 2: Simplify further using the trigonometric identity sin(A+B) = sinAcosB + cosAsinB.
[9sinθ+2sinθcos(π/3)+2cosθsin(π/3)]^2 = [9sinθ+(2/2)sinθcos(π/3)+(2/2)cosθsin(π/3)]^2
Step 3: Simplify the coefficients.
[9sinθ+(1)sinθcos(π/3)+(1)cosθsin(π/3)]^2 = [10sinθ+(cosθ)(sin(π/3))]^2
Step 4: Use the trigonometric identity sin(π/3) = √3/2.
[10sinθ+(cosθ)(√3/2)]^2
Step 5: Find the maximum value of the expression.
To find the maximum value of the expression, we need to find the maximum value of each term inside the square bracket.
Given that |sinθ| ≤ 1 and |cosθ| ≤ 1, the maximum value of the expression would occur when sinθ = 1 and cosθ = 1 (both at their maximum values).
Substituting sinθ = 1 and cosθ = 1 in the expression, we get:
[10(1)+(1)(√3/2)]^2 = [10+√3/2]^2
Simplifying further, we get:
[10+√3/2]^2 = (20+√3)^2
Therefore, the maximum value of the expression [9sinθ+2sin(θ+π/3)]^2 is (20+√3)^2.
To find the maximum value of the expression [9sinθ+2sin(θ+π/3)]^2, we can start by expanding and simplifying the expression.
Let's expand the square using the formula (a+b)^2 = a^2 + 2ab + b^2:
[9sinθ+2sin(θ+π/3)]^2 = (9sinθ)^2 + 2(9sinθ)(2sin(θ+π/3)) + (2sin(θ+π/3))^2
Simplifying each term individually, we get:
(81sin^2θ) + 2(18sinθsin(θ+π/3)) + (4sin^2(θ+π/3))
Now we can simplify further using trigonometric identities. Recall the identities:
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
sin^2(a) = 1 - cos^2(a)
Start by simplifying the middle term of the expanded expression, 2(18sinθsin(θ+π/3)):
Multiply the coefficients and apply the trigonometric identity:
2(18sinθsin(θ+π/3)) = 36sinθsin(θ+π/3)
= 36sinθ(cos(π/3)sinθ + sin(π/3)cosθ)
= 36sinθ(sinθ/2 + (√3/2)cosθ)
= 18sin^2θ + 18√3sinθcosθ
Now, simplify the last term of the expanded expression, (4sin^2(θ+π/3)):
Apply the trigonometric identity:
4sin^2(θ+π/3) = 4[1 - cos^2(θ+π/3)]
= 4 - 4cos^2(θ+π/3)
Now we can rewrite the expanded expression using the simplified terms:
(81sin^2θ) + 18sin^2θ + 18√3sinθcosθ + 4 - 4cos^2(θ+π/3)
Combine like terms:
(99sin^2θ) + 18√3sinθcosθ + 4 - 4cos^2(θ+π/3)
To find the maximum value of this expression, we can recognize that sin^2θ and cos^2θ have maximum values of 1. This means that the terms (99sin^2θ) and 4 will always be positive and contribute to the maximum value.
For the term 18√3sinθcosθ, we can use the fact that its maximum value occurs when sinθ and cosθ are both at their maximum (which is 1/sqrt(2)). Thus, this term contributes a positive value to the maximum.
Lastly, the term -4cos^2(θ+π/3) has a maximum value of -4 since cos^2(θ+π/3) has a maximum value of 0.
Summing up all the terms, the maximum value of the expression [9sinθ+2sin(θ+π/3)]^2 is:
(99sin^2θ) + 18√3sinθcosθ + 4 - 4cos^2(θ+π/3)
= 99(1) + 18√3(1/sqrt(2))(1/sqrt(2)) + 4 - 4(0)
= 99 + 18(3/2) + 4
= 99 + 27 + 4
= 130.
Therefore, the maximum value of [9sinθ+2sin(θ+π/3)]^2 is 130.