Solving a Trig Equation Using Sine of a SumDate: 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/ |
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