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# 1. Let be an angle. Then there exists constants a and b such that for all . Enter a, b in that order.

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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
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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   Sep 19, 2019
#2
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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   Sep 19, 2019