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So my buddy and I (both HS Math teachers) have been messing around with a question about finding all solutions of a "co-function" equation like the one above. The typical HS questions asks students to find the solution in Q1 by solving $2x + 3x = 90$.

He was thinking about how to get the other solutions and we were curious to see what other people do for a problem like this.

We've noticed that the solutions come in regularly spaced intervals from two separate starting points.

Namely, $\frac{\pi}{10} \pm \frac{4k\pi}{10}$, AND $-\frac{\pi}{2} \pm 2k\pi$ {$k$ natural} for the above equation.

We've been able to generalize this to any such co-function equation of the form $\sin(ax)=\cos(bx)$ {$a,b$ Integers} , and are curious about whether other people have stumbled upon this as well.

We've also discovered a very interesting geometric interpretation of the periodic solutions and where they occur.

We'd love to hear how you approach this type of problem (algebraically, not graphically) and whether you've encountered these ideas as well.

N. F. Taussig
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Mike Andrejkovics
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    Curious, in your generalization does it matter that a<b? – MathHype Mar 26 '15 at 23:58
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    Yes, it affects the sign of the starting position of the second set of solutions. So, if a<b, then you will have solutions at: -pi/2(b-a) plus or minus 2pi/(b-a), and if a>b solutions at: pi/2(a-b) plus or minus 2pi/(a-b). Either way, you also have a set of solutions at: pi/2(a+b) plus or minus 2pi/(a+b) – Mike Andrejkovics Mar 27 '15 at 00:16

4 Answers4

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$\sin(2x) = \cos(3x) \Rightarrow \cos(\frac{\pi}{2}-2x) = \cos (3x) \Rightarrow \dfrac{\pi}{2} - 2x = \pm 3x + 2k\pi, k \in \mathbb{Z}$. Can you take it from here?

DeepSea
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  • Nice! That's a compact and efficient way to handle it : ) – Mike Andrejkovics Mar 27 '15 at 00:29
  • I think, to be rigorous, the same expansion should be used for both cosines, so it becomes $\pm(\pi/2-2x)+ 2j\pi=\pm 3x +2k\pi, k, j \in \mathbb{Z}$ where all possible choices of "$\pm$" can be used. – marty cohen Mar 27 '15 at 02:19
  • @martycohen I think the way it is is more elegant; we have something of the form $f(x)=f(y)$ and rather than apply the preimage $f^{-1}$ to both sides as you suggest, it's faster to use the equivalent $y\in f^{-1}[{f(x)}]$ which is much faster. It happens that all preimages are of the form $\pm c + 2k\pi$. – Milo Brandt Mar 27 '15 at 03:07
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There is another way. You can use the sine and cosine addition formulas to recursively factor out the whole thing. It takes a lot of patience, but at the end you get a quadratic equation you can solve.

$\sin (2x)=\cos(3x)$

$2\sin(x)\cos(x)=\cos(2x+x)$

$2\sin(x)\cos(x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)$

$2\sin(x)\cos(x)=(\cos^2(x)-\sin^2(x))\cos(x)-2\sin^2(x)\cos(x)$

Then you divide everything by $cos(x)$ giving you...

$2\sin(x)=(\cos^2(x)-\sin^2(x))-2\sin^2(x)$

Use the identity $\cos^2(x)=1-\sin^2(x)$

$2\sin(x)=1-4\sin^2(x)$

Now you have something that you can substitute for to solve a quadratic. For example say $w=\sin(x)$

Then your equation looks like $2w=1-4w^2$ which is really $4w^2+2w-1=0$ in disguise. Solve the rest on your own.

MathHype
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One way is to use the identity

$$\sin \cos^{-1} = \sqrt{1 - x^2}$$

after making some substitutions. There's a family of similar identities for other trig functions.

Qudit
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$$ \sin 2x = \sin ( \pi/2 -3 x) $$ $$ 2 x = \pi/2 - 3 x, 2 x = \pi - \pi/2 + 3 x $$ $$ x = \pi/10, -\pi/2 $$

and all co-terminal angles.

Narasimham
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  • but not co-terminal to those x-values... we're looking for the x-values that generate angles co-terminal to the 2x and 3x above- yes? – Mike Andrejkovics Mar 27 '15 at 00:34
  • @MikeAndrejkovics: on the second line you add $2k\pi$ to generate all the solutions and then solve for $x$. That will sort out the periodicities. – Zorawar Mar 27 '15 at 01:44
  • Right. The question poser was long , so answer is kept simple. – Narasimham Mar 27 '15 at 05:49