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# Help quick

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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
+109740
+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

Nov 5, 2019
edited by CPhill  Nov 5, 2019
edited by CPhill  Nov 6, 2019
#2
+2850
+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
+109740
0

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

CPhill  Nov 6, 2019
#4
+1

sorry cphill but its $$\frac{\sqrt{35}}{3}$$

Guest Nov 6, 2019
#5
+109740
+1

I just rounded

[√105 - 1 ]

________    =    sin  θ

12

So

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

√35

___   ≈   1.972

3

CPhill  Nov 6, 2019