Proving equivalence of two sinusoidal formulas

[xodin]

Homework Statement

While doing a physics assignment today, I came up with an expression in my answer that appears to be equivalent to sin(θ), however, i couldn't find a way to manipulate the expression to prove that it was. Could anyone point me in the right direction as to a way that I could have shown that these two expressions are equal, without having to use the support of a logical argument or graph? I want to figure out how I could have done this using nothing but algebra and trig identities.

Homework Equations

\sqrt{\frac{37}{4} + 3sin(θ)} - \sqrt{\frac{37}{4} - 3sin(θ)} = sin(θ)

The Attempt at a Solution

I've tried what I can but don't get very far. I even enlisted the help of a computer algebra system to see if it simplified it in a way I didn't recognize, but no joy. BTW, this is just for curiosity, not actual homework. Thanks for any help provided!

 

[Dick]

Homework Statement

While doing a physics assignment today, I came up with an expression in my answer that is equivalent to sin(θ), however, i couldn't find a way to manipulate the expression to prove that it was. Could anyone point me in the right direction as to a way that I could have shown that these two expressions are equal, without having to use the support of a logical argument or graph? I want to figure out how I could have done this using nothing but algebra and trig identities.

Homework Equations

\sqrt{\frac{37}{4} + 3sin(θ)} - \sqrt{\frac{37}{4} - 3sin(θ)} = sin(θ)

The Attempt at a Solution

I've tried what I can but don't get very far. I even enlisted the help of a computer algebra system to see if it simplified it in a way I didn't recognize, but no joy. BTW, this is just for curiosity, not actual homework. Thanks for any help provided!

It's not really correct. The left side is actually a pretty good approximation to the right side, but it's not exact. Take your calculator and put theta=1.

 

[xodin]

It's not really correct. The left side is actually a pretty good approximation to the right side, but it's not exact. Take your calculator and put theta=1.

Oh, that's interesting. Thanks for the response. The graphs appear to overlap exactly, but of course there could be very minor differences that the resolution can't resolve. Are you positive it's the expressions that aren't equal and not the calculator's approximation of floating point values that's causing the disparity?

 

[xodin]

Oh, that's interesting. Thanks for the response. The graphs appear to overlap exactly, but of course there could be very minor differences that the resolution can't resolve. Are you positive it's the expressions that aren't equal and not the calculator's approximation of floating point values that's causing the disparity?

Actually I was able to use a different graphing application and turned the resolution way up by zooming in, and you're right, they are just a hair different. Amazing. Thanks for your help!

EDIT: Turns out they are only ~0.0035 apart from each other at the widest spot of divergence (near 0.5 and -0.5) and virtually inseparable near 0, 1, and -1.

 

[Dick]

Actually I was able to use a different graphing application and turned the resolution way up by zooming in, and you're right, they are just a hair different. Amazing. Thanks for your help!

EDIT: Turns out they are only ~0.0035 apart from each other at the widest spot of divergence (near 0.5 and -0.5) and virtually inseparable near 0, 1, and -1.

In fact they are exactly equal when sin(θ)=0, 1 and -1. Now you've got me all curious. How did you come up with that contraption?

 

[xodin]

In fact they are exactly equal when sin(θ)=0, 1 and -1. Now you've got me all curious. How did you come up with that contraption?

I was working a physics problem with two radio beacons spaced d=20 m away, each +/- d/2 away from the origin in the y-direction. They were broadcasting in-phase waves of 0.5 m wavelength, and there was a plane flying in a radius r=3*d=60 m circle with a radio that was cutting in and out due to the intermittent interference. We had to find the path length difference as a function of θ, phase difference as a function of θ, and number of maxima (max constructive interference angles) that occur in one trip around the circle. Similar to two-slit interference problems.

Basically what happened was I thought I could use Young's two-slit formula to solve this, which is ΔL = d sinθ = mλ, but the instructions said I couldn't. So instead I used the vectorial path length difference, which was:

ΔL = d(\sqrt{\frac{37}{4} + 3 sin(θ)} - \sqrt{\frac{37}{4} - 3 sin(θ)}) = mλ

However, when solving for possible values of m, I tried it out both ways and found they had the same result, so I opened up my grapher application and graphed them both and they are remarkably identical; but you are right that they do diverge very, very slightly--roughly 0.0035 or so y-value difference at the most divergent spot.

 

[haruspex]

Intriguing.
If you write √(α+ n x) - √(α- n x) = x then this simplifies to x=0 or α = n² + x²/4. With α = n² + 1/4, it's a perfect match at x = 0 and ±1.

 

[xodin]

Just for fun I made some measurements using a computer program. The maximum difference between the values of the two functions at any given point is 0.005321, and it occurs four times, at pi/5, 4pi/5, 6pi/5, and 9pi/5.

 

[Chestermiller]

Substitute sinθ = y, and then expand the left hand expression in a McLauren series in y. See what you get. The y² term will drop out, and the coefficient of y will be √(36/37).

 

[Chestermiller]

Oop. I forgot this is precalculus math. Instead of expanding in a McLauran series, express each of the terms on the left hand side using the binomial theorem.