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1. Let \(\theta\) be an angle. Then there exists constants a and b such that \(\cos\left(\theta + 60^{\circ}\right) = a \sin (\theta) +b \cos (\theta)\) for all \(\theta\). Enter a, b in that order.

 

2. Let \(\theta\) be an angle. Then there exist constants a and b such that \(\sin\left(\theta + \arctan\left(\frac{5}{12} \right)\right) = a \sin (\theta) +b \cos (\theta)\) for all \(\theta\). Enter a, b in that order.

 

3. Find the greatest possible value of \(4 \sin (\theta) + 3 \cos (\theta)\).

 

Thanks for the help!

 Sep 19, 2019
 #1
avatar+106535 
+1

We are using the angle sum formulas here :

 

1)   cos ( θ + 60°)  =  cos (60°) sin (θ)  -  sin(60°)cos(θ)  = (1/2)sin(θ) - (√3/2)cos(θ)

 

So     a  =  1/2      and   b   =  -√3/2

 

 

2)   sin ( θ  + arctan (5/12) )   =   cos (arctan (5/12)) * sin(θ ) + sin (arctan (5/12)) * cos (θ) 

 

Note 

cos ( arctan (5/12) )  =   12/sqrt (5^2 + 12^2)  =  12/sqrt(169)  = 12/13   = a

sin (arctan (5/12) )  =  5 / sqrt (5^2 + 12^2)  = 5/sqrt (169)  =  5/13   = b

 

 

cool cool cool

 Sep 19, 2019
 #2
avatar+106535 
+1

3)    4sin (θ)  + 3cos(θ)

 

Take the derivative and set to  0

 

4cos (θ)  - 3sin(θ)  = 0

 

4cos (θ)  = 3 sin (θ)

 

(4/3)  =  sin (θ)/cos(θ)

 

(4/3)  = tan (θ)

 

arctan (4/3)  = θ

 

So

 

4 sin ( arctan (4/3))  + 3cos (arctan (4/3)  =

 

4 (4 / sqrt (4^2 + 3^2))  + 3 ( 3/sqrt (4^2 + 3^2) ) =

 

16 / sqrt (25)  +  9 /sqrt (25)  =

 

16/5 +  9/5  =

 

25/5  =

 

5

 

 

cool cool cool

 Sep 19, 2019

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