prove that mg sin theta is equal to mg tan theta when sin theta is almost equal to theta in radians

Question

It's about an experiment that asks you to determine the free-fall acceleration, g , using simple pendulum

Answer 1

I like Arturo O's answer.    I am going to propose another one.   We want to demonstrate that for small angles  sin(x) = tan(x)

 

To show that to be true let's look at the ratio of  sin(x)/cos(x) as x approaches 0 (small angle approximation)

 

lim  sin(x)/tan(x) =  lim [sin(x) / (sin(x)/cos(x))] = lim x->0  cos(x) = 1;

 

So for small angles  sin(x)/tan(x) is nearly 1, and sin(x) is nearly equal to tan(x)

Answer 2

That is equivalent to proving that sinθ ≅ tanθ, which is true only for θ << 1 radian.  (It is not true in general.)  To see that this holds for very small θ, you could look at the power series for sin(x).

 

sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + ...

 

For very small x, drop terms of order x² or higher.  Then

 

sin(x) ≅ x for x << 1

 

Similarly,

 

cos(x) = √(1 - sin²x) ≅ √(1 - x²) for very small x.  You can use the approximation

 

√(1 - x²) ≅ 1 - (1/2)x²,

 

which is valid for very small x.  For very small x, drop terms of order x² and above.  Then

 

cos(x) ≅ 1 for x << 1

 

You are left with

 

tan(x) = sin(x)/cos(x) ≅ x/1 = x

 

tan(x) ≅ sin(x) ≅ x for very small x.  This is the "linearizing approximation" used frequently in dynamics problems when the angles are very small compared to 1 radian.