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If \(0 < \theta < \frac{\pi}{2}\) and \(\sqrt{3} \cos \theta - \sin \theta = \frac{1}{3},\) then find \(\sqrt{3} \sin \theta + \cos \theta.\)

 Nov 5, 2019
 #1
avatar+106533 
+2

√3 cos θ  - sin θ  =  1/3

 

 

 √3 cos θ  =  sin θ  +1/3       square both sides

 

3cos^2θ  =  sin^2θ + (2/3)sin θ + 1/9

 

3 (1 - sin^2θ)  = sin^2 θ + (2/3)sin θ  + 1/9 

 

3 - 3sin^2 θ = sin^2θ + (2/3)sin θ + 1/9         rearrange as

 

4sin^2 θ + (2/3) sinθ - 26/9  =  0             multiply through by 9

 

36sin^2θ +6sin θ - 26  =   0     

 

18 sin^2 θ + 3sin θ - 13  = 0          let   sin θ  =  x

 

18x^2  + 3x  -  13  =  0

 

Using the quadratic formula  and taking the poisitve result  [ since the sine is positive here, we get that ]

 

x  = sin θ  =  - 3   + √[ 3^2 + 4* 18 * 13]

                    ______________________  ≈  .77

                              2 * 18

 

 

Using the sine inverse  we get that 

 

arcsin (.77)  = θ  ≈  .88 rads

 

So

 

3 sin (.88)  + cos (.88)  ≈  1.972

 

 

 

cool cool cool

 Nov 5, 2019
edited by CPhill  Nov 5, 2019
edited by CPhill  Nov 6, 2019
 #2
avatar+2551 
+1

I do not understand any of those trig gibberish yet, but nice job Cphill! One day I will learn the dark secrets of sine cosine and tangent!

CalculatorUser  Nov 6, 2019
edited by CalculatorUser  Nov 6, 2019
 #3
avatar+106533 
0

Don't worry, CU......as good as you are in math, the trig "secrets"  will come easily to you....!!!!

 

 

 

cool cool cool

CPhill  Nov 6, 2019
 #4
avatar
+1

sorry cphill but its \(\frac{\sqrt{35}}{3}\)

Guest Nov 6, 2019
 #5
avatar+106533 
+1

I just rounded

 

The quadratic  evaluates to

 

[√105 - 1 ]

________    =    sin  θ

     12

 

 

So

 

√3sin  ( arcsin ( [√105 - 1 ]/12 ) )  +  cos (arcsin ( [√105 - 1 ] /12) )  =

 

√35

 ___   ≈   1.972

   3

 

 

cool cool cool

CPhill  Nov 6, 2019

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