# Linear Combinations of Trig Functions - Finding Roots

1. Mar 11, 2010

### david_p

Hi there
I was wondering if there is a simple way to solve for the roots of a complicated summation of trig functions that can't be combined with any simple identities.
I have an equation of the form:

0 = sin(8x-arctan(4/3))+3.2sin(16x+pi/2)

where the two sines have different amplitudes, phases and a frequency that differs by a factor of 2. I would like to find the values of x on [0,pi/2] that satisfy this condition. I don't know very much about Fourier analysis but is there some method for solving this?
I need an analytical solution - not an approximation

2. Mar 11, 2010

### Pengwuino

I don't think you need to go into anything complicated. What does $$Sin(\theta + \frac{\pi}{2})$$ equal?

3. Mar 11, 2010

### david_p

Yes, it equals cosx but the frequency is still different so you can't easily combine the two terms. I dont think there is any way of combining the two terms into one simple function. The solution must have an amplitude that oscillates as a function of x; like a form:
Asin(Bx+C)sin(Dx+E)

Not sure how to solve for the parameters though

4. Mar 11, 2010

### Pengwuino

Oh right, I didn't see that, maybe this isn't as simple as it looks. I'm assuming you've exausted double angle and angle addition formulas?

5. Mar 11, 2010

### david_p

Yes, Ive tried almost every standard substitution. I dont think this will solve the problem since the amplitude varies with x. Plotting the function in maple reveals that it doesn't have the form of a simple cosine function

6. Mar 12, 2010

### uq_civediv

you could still use the cosx if you treat it as sin(16x+pi/2) = cos(16x) = cos(2*8x) = 2cos^2(8x)-1
whereas you could expand the first term sin(8x- ... getting it in terms of sin8x and cos8x, substituting sin8x=sqrt(1-cos^2(8x)) and squaring to eliminate the sqrt leaves you with a 4th degree polynomial of cos8x. which is technically solvable

if i thought it through enough, didn't try in written

7. Mar 12, 2010

### david_p

No this wont work either, you will end up with a function that has non integer cos terms
Unless I misread what you were saying

8. Mar 12, 2010

### Redbelly98

Staff Emeritus

9. Mar 12, 2010

### david_p

Its an assignment question,
but i'm also curious if there is a generalized technique for approaching this type of problem

10. Mar 12, 2010

### uq_civediv

you don't need integer terms to solve a 4th degree polynomial, though you can just multiply the eq with sth to make them integer

$$Sin(8x-arctan(4/3))+3.2Sin(16x+\pi/2)= \pm 0.6 \sqrt{1-Cos^2(8x)}-0.8Cos(8x)+6.4Cos^2(8x)-3.2 = 0$$

or
$$\pm 3 \sqrt{1-a^2}-4 a+32a^2-16=0$$
where a=cos(8x)

Last edited: Mar 12, 2010