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Solving a Trig Equation Using Sine of a Sum

Date: 08/20/2008 at 13:43:48
From: Janice
Subject: Solve 4 cos x + 6 sin x = 5

Solve 4 cos x + 6 sin x = 5.

I don't know where to start.  I've looked at similar questions and
solutions from Dr. Math's archives.  I've tried solving this question
by identifying necessary identities I need to use.

I believe one of the identities is sin (A+B) = sinAcosB + cosAsinB

Thanks.


Date: 08/20/2008 at 18:37:41
From: Doctor Anthony
Subject: Re: Solve 4 cos x + 6 sin x = 5

Hi Janice,

Thanks for writing to Dr. Math.  You have:

  4 cos(x) + 6 sin(x) = 5

Consider the identity for the sine of a sum, and multiply it by a
constant R:

  R*sin(@ + x) = R*[sin@cos(x) + cos@sin(x)]

               = Rsin@cos(x) + Rcos@sin(x)

Compare that result with your left side of 4 cos(x) + 6 sin(x).  To
make them match, we let Rsin@ = 4 and Rcos@ = 6.  We can use those
equations to solve for R:

  Rsin@ = 4  -->  R^2sin^2(@) = 16
  Rcos@ = 6  -->  R^2cos^2(@) = 36
                 -----------------
                  R^2sin^2(@) + R^2cos^2(@) = 16 + 36
                  R^2(sin^2(@) + cos^2(@))  = 52
                                        R^2 = 52
                                          R = 2sqrt(13)

Dividing the two equations we can solve for @:

  Rsin@ = 4
  ---------  -->  tan@ = 4/6 = 2/3  -->  @ = 33.7 degrees
  Rcos@ = 6    
  
And so        Rsin(x+@) = 5   becomes

  2sqrt(13)*sin(x+33.7) = 5

            sin(x+33.7) = 0.693375
               x + 33.7 = 43.898
                      x = 10.198 degrees

Check:  4 cos(10.198) + 6 sin(10.198) =  4.999   (so near enough)

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 08/21/2008 at 15:27:49
From: Janice
Subject: Thank you (Solve 4 cos x + 6 sin x = 5)

Hi Doctor Anthony,

Thanks for your help.  At first, I was confused about the solution,
but I've taken time to work it out.

Thanks again.



Date: 08/21/2008 at 21:45:00
From: Doctor Ali
Subject: Re: Solve 4 cos x + 6 sin x = 5

Hi Janice!

Thanks for writing to Dr. Math.  To add to Dr. Anthony's work, here's
a slightly different way to solve the equation.

The first thing to note about

  a sin(x) + b cos(x) = c

is that the equation may have no solutions.  To check that, we start 
by dividing both sides by sqrt(a^2 + b^2). 

That gives us

        a                       b                      c
  -------------- Sin(x) + ------------- Cos(x) = -------------
  Sqrt(a^2+b^2)           Sqrt(a^2+b^2)          Sqrt(a^2+b^2)

The above expression can be written as the sine of the sum of two
angles.  That is, there exists an angle y such that

                b
  Sin(y) = --------------
           Sqrt(a^2+b^2)

                a
  Cos(y) = --------------
           Sqrt(a^2+b^2)

Divide the above equalities to get

            b
  Tan(y) = ---
            a

You can find y from here,

            b
  y = Atan(---)
            a

Now, we can change the equation to,

                 b                c
  Sin( x + Atan(---) )  =  ---------------
                 a         Sqrt(a^2 + b^2)

As you see, the equation is now changed to a familiar form.  Note 
that if,

  |        c        |
  | --------------- | > 1
  | Sqrt(a^2 + b^2) |

the equation has no real solution.  Does that make sense?

See also this answer from our archives:

  Obliterating Iterating (Fazed by Phase)
    http://mathforum.org/library/drmath/view/65138.html 

Please write back if you still have any difficulties.

- Doctor Ali, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Trigonometry

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