Proving the ratio of Sine and respective angle as unit value i.e 1 mathematically is objective in this article. This limits’ function mostly helps in dealing differentiation. In other words, it is most important formula in calculus.
In order to prove it mathematically, consider some assumptions. Assume θ is an angle of right angle triangle and with respect to angle the trigonometric function Sine is Sin θ. The angle of right angle triangle is not zero but approximately zero which means θ → 0 or θ ≈ 0.
The ratio of Sine and its angle is (Sin θ)/θ. By applying the limits function with angle’s assumption, it can be expressed as written below mathematically.
| lim θ → 0 |
θ |
We have already known that, the angle of right angle triangle θ → 0. The angle is not zero truly but it is approximately zero in value. So, let us check sine value function with some values which are approximately zero in value.
I got these rules with the help of my scientific calculator in radians.
Sin 0.000056 = 0.0000559999 ≈ 0.000056
Sin 0.000235 = 0.0002349999 ≈ 0.000235
Sin 0.0045 = 0.0044999848 ≈ 0.0045
Sin 0.089546 = 0.089426377 ≈ 0.089546
Sin 0.256489 = 0.253685979 ≈ 0.256489
If you observe these values, you can clearly understand that for the smaller angles, trigonometric ratio sine gives some values which are approximately equal to the respective angles. In other words, we can understand that Sin θ ≈ θ if the angle is very small in value.
Substitute θ instead of Sin θ in the limits function.
| lim θ → 0 |
θ |
≈ |
lim θ → 0 |
θ |
The value of numerator and denominator is equal and they both get cancelled.
| lim θ → 0 |
θ |
There is no θ term in function. So, the value of this function is unit value i.e 1.
| lim θ → 0 |
| lim θ → 0 |
θ |
1 |
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