# Complicated trigonometric equation

Chen
I need to solve
sin(ax)sin(bx) - k cos(ax)cos(bx) = -1
where k > 1 is constant and so are a and b, and they are all irrational.

(It's part of a much larger question in optics...)

Thanks

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Homework Helper
Gold Member
Dearly Missed
Well, you could rewrite the left-hand side as:
$$\frac{1}{2}(\cos(ax-bx)-\cos(ax+bx))-\frac{k}{2}(\cos(ax+bx)+\cos(ax-bx))=-1$$
Which can be simplified to:
$$(1-k)\cos(ax-bx)-(1+k)\cos(ax+bx)=-2$$
which is some simplification at least..

tim_lou
if all things fail... maybe you can use complex numbers.

Chen
Thanks for your tips. I was able to get the equation to the form:
$$a^2 cos(ax) + b^2 cos(bx) = c$$
where a,b,c are different constants, but still irrational. Any ideas from here? :)

Thanks!

Swapnil
Thanks for your tips. I was able to get the equation to the form:
$$a^2 cos(ax) + b^2 cos(bx) = c$$
where a,b,c are different constants, but still irrational. Any ideas from here? :)

Thanks!

I spent half-an-hour trying to solve it for x. But I couldn't. I am starting to think that there is no closed form solution to this equation. Anyone else share the same thought?

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Homework Helper
I don't know if im right, lets just go with this.
For the original equation:$$\sin (ax)\cdot\sin (bx) - k\cdot \cos (ax) \cos (bx) =-1$$ Let $$-1=-\sin^2 (ax) -\cos^2 (ax)$$ and see what happens.

morson
I really have no clue. I tried to simplify it a bit:

sin(ax)sin(bx) - kcos(ax)cos(bx) + 1 = 0

sin(ax) [ sin(bx) - {kcos(ax)cos(bx)}/sin(ax) + 1/sin(ax) ] = 0

sin(bx) - kcsc(ax)cos(bx) + cot(ax) = 0

At this point I'm stumped. I fiddled around and got something equally unhelpful:

tan(bx) + cot(ax)sec(ax) = kcsc(ax)

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Boorglar
I think I found a way to do it...

I am frustrated because I typed my whole solution here and then when I sent it it turned out my account was disconnected while I was writing. So I won't bother writing it out again in detail.

But the general idea is to start with the equation given by arildno and multiply it by -1 (to get rid of the minus signs).

Then use this formula with I am not quite sure if it works in all cases:

A cos α + B cos β =
2 cos{[arccos(A cos α) + arccos(B cos β)]/2} cos{[arccos(A cos α) - arccos(B cos β)]/2}

Where A = (k-1) , α = (a-b)x , B = (k+1) and β = (a+b)x.

Divide by 2 on both sides to get rid of the 2's.

Then you have a product of cosines which equals 1. So each cosine equals 1 or -1 as well.

This gives you a system of equations.

You can solve this system easily (just take care of the range of arccos). One problem is that in the end, you'll have two expressions for x, both of which must be equal, and so you'll have to choose the appropriate integers in the + 2n$\pi$ and + 2m$\pi$ (if they even exist, I have no idea how to find them besides trial and error).

The other problem is that this formula doesn't seem to work for k between -2 and 2 because of the domain of arcsin(1/(k-1)) (you'll have to solve the system to see what I mean).

EDIT:

The final answer I have is
x = 1/(a-b) *[±arcsec(k-1) + 2n$\pi$]
x = 1/(a+b)*[±arcsec(k+1) + 2m$\pi$]

(x satisfies both conditions, so not all n and m will actually work)

A second solution is almost the same but with (1-k) and (-k-1) as the argument in arcsec.

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mathdude1998
Hi I don't know if this will help but i calculated the factorising of it and the answer is abs2x2(-1c2ko2 + i2n2) . Hope it helps.