# Mathematical approximations in Physics for angles <20 degrees

1. Aug 16, 2011

### hyde2042

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

62. In physics, it is important to use mathematical approximations.
(a) Demonstrate that for small angles (<20°)

tan a ~ sin a ~ a = (Pia'/180 degrees)

where a is in radians and a' is in degrees. (b) Use a calculator
to find the largest angle for which tan a may be
approximated by a with an error less than 10.0%.

2. Relevant equations

3. The attempt at a solution

I'm unsure how I would go about this other than plugging in numbers less that 20 degrees, but I wouldn't know what is within ten percent error.

2. Aug 16, 2011

### PeterO

If you take sin 30 you get 0.5000. If you take tan 30 you get 0.5773

That is a difference of 0.0773

That represents a percentage error of (0.0773/.5000 * 100)% or 15.5% difference.
That is bigger than 10%.

If the angle is smaller , the error is smaller.
NOTE: error doesn't mean a mistake, it means how far from the real answer would you calculated answer be if you used tan rather than sin - perhaps for some convenience reason.

3. Aug 16, 2011

### hyde2042

Oh, as you go from tan (a) to sin (a) to the actual (a) in radians it will be equal to the angle of (a) times Pi over 180?

Edit: Thanks for your time and reply. It feels embarrassing that I should be strugglling with these questions in the first chapter as opposed to all the other questions I've been seeing.

4. Aug 16, 2011

### PeterO

the "angle of (a) times Pi over 180" is just there to convert degrees to radians

5. Aug 16, 2011

### hyde2042

Ah I see, thank you. It's been a few years since my Trig class.

6. Jun 20, 2013

### mbyoung18

I understand that tan of a small angle is close to the sine of an angle, but what is the significance of it.

In my book it states for waves that Tsin(theta) for small angles is approximately equal to Ttan(theta) and we should use that instead. My question is why? Why use tangent instead of sine?

Thanks

7. Jun 20, 2013

### PeterO

If you are looking at an interference pattern on a screen for example.

The source of the pattern (The actual double or single slit) is, say, 1 m from a screen.
If the feature (max or min) you are working with it 2cm off centre, then the angle to that feature has a tangent of 2/100.
If there is another feature 3cm off centre, then the angle to that feature has a tangent of 3/100.

If you were after the sine of each of those angles you would have to measure the direct distance to each feature in turn. (tan is opposite/adjacent; sin is opposite/hypoteneuse)

If the sin and tan value are basically the same, you are saved that problem.

You would also be saved the problem if the subsequent formulae you were using had the tan function in them, but I think the formulae include sine.