# Trig Identities

1. Jan 9, 2008

### mrroboto

1. The problem statement, all variables and given/known data

4.59 x 10^-4 = Sin^3(x)/Cos(x)

Solve for X

2. Relevant equations

Sin(x)/Cos(x) = Tan(x)

3. The attempt at a solution

We can make Sin^3(x)/Cos(x) into Tan(x)Sin^2(x), but I don't think that helps...

What trick do I use?

Last edited: Jan 10, 2008
2. Jan 11, 2008

### Gib Z

I can't think of an exact solution right now, but if we just use small angle approximations of sin and cos backwards, we can get a VERY close solution !!

Just take cube roots of both sides,

$$9.6702579 \cdot 10^{-11} = \frac{\sin x}{( \cos x)^{1/3} }$$.

With our backwards small angle approximation attempt, we find x = $9.6702579 \cdot 10^{-11}$.

When we sub that back in for x on the right hand side on your original expression, we get approximately:

0.000458999999999999999999999999999999999999999554012791.......

Which means we have an error of $$4.45987209 \cdot 10^{-46}$$.

I think thats quite reasonable, dont you?

3. Jan 11, 2008

### mda

This is along the right track (although there is a non-trivial analytic solution)...
but the numerical answer isn't quite right...

4. Jan 11, 2008

### Gib Z

Can you show me the analytic solution? And What I know the answer its quite right, It is reasonably accurate though, no? Of course, we can make it more accurate by taking more terms of the taylor series and with some error analysis, solving quartic equations to arbitrary accuracy with newtons method - That all takes so much time for so little extra result.

5. Jan 11, 2008

### TheoMcCloskey

GibZ - you cubed the constant instead of taking the cube root. I think the answer is more closer to x = 0.079

6. Jan 11, 2008

### mda

GibZ, I won't show the solution I got from Maple because it is absolutely horrific, would not teach us anything and as you say is a waste of time anyway. :)
I see Theo has posted the right answer so I'll leave it at that.

7. Jan 11, 2008

### TheoMcCloskey

Ya know, you can re-write the original expression in terms of, say, $u = {\cos}^2(x)$. This will be a cubic polynomial in u. You could then search for the zero of this polynomial. Of course, I would still do so numerically as you have available a very good initial estimate.

So, questions for you:

1. how would you formulate the polynomial P(u). (Hint: convert every thing in terms of cos(x)).

2. what is this initial estimate I speak of. (Hint: expand the numerator and denominator of the quotient $\tfrac{{\sin}^3(x)}{\cos(x)}$)

Of course, all of this is for small x solutions. Are there any others? I think maybe the answer is Yes

8. Jan 12, 2008

### Gib Z

DAMN IT.

Well, point is - Use small angle approximations for this. Answer should still be very accurate.

9. Jan 12, 2008

### dynamicsolo

By any chance, did this equation come up from working on the physics problem involving two hanging, charged pith balls, where you have to solve for the angle the cords make to the vertical at equilibrium? (Having worked with students on this problem multiple times, I've gotten used to seeing this expression...)

As has already been pointed out, if the product [ tan(x) · sin^2(x) ] is much smaller than 1, you can safely use the small angle approximation for sine and tangent to get a good first estimate for the solution. With sin(x) and tan(x) approximately equal to x in radians, you can approximate your equation by

x^3 = 4.59 x 10^-4 ,

which gives you a first guess of x = 0.0772 radian. You can then put this into your exact product, [ tan(x) · sin^2(x) ] , and see what you get. Since sine and tangent will both increase with increasing x for the angles you'd be working with, you can then "tweak" your estimate for x up or down to move the product up or down. [In this case, you'd need to lower the value a touch.] For most of the problems of this sort I've seen, you can usually get to three sig-figs of precision in three or four passes... (Make sure, naturally, that your calculator is in radian mode when doing this.)

Then, of course, you could also solve this graphically...

Actually, 0.079 is a little high. The cube-root estimate turns out to be very close in this case.

Last edited: Jan 12, 2008
10. Jan 12, 2008

### Gib Z

If you have said the first sentence to me in real life, without the second, I would be terrified =] Just in case this is more of a maths problem than physics, writing the expression in terms of sin and tan makes the period, and hence the general form of the solution, more obvious.