# Sin(theta) = dy/dx ?

In preparing for an acoustics course, I ran across the following sentence which confused me:

"If (theta) is small, sin(theta) may be replaced by [partial]dy/dx."

I expected to see sin(theta) = (theta) so this threw me off. This came up in the derevation of the one dimensional wave equation after approximating (by Taylor series) the transverse force on a mass element of a tensioned string with [partial]d(Tsin(theta))/dx. The approximation in question thus gave T*([partial]d2y/dx2)*dx.

In the original setup, x and y are cartesian axis in physical 2D space and (theta) is the angle the string (with tension T) makes from the x-axis after displacement from equalibrium.

I've never seen sine approximated by dy/dx before and was hoping somebody might shed some light for me :)

LCKurtz
Homework Helper
Gold Member
When ##\theta## is small, ##sin\theta \approx \tan\theta = \frac{dy}{dx}##. In the derivation for the vibrating string, ##\theta## is the slope angle of the string.

Thank you! :)

rcgldr
Homework Helper
sin(θ) = Δy / sqrt(Δy^2 + Δx^2). For small θ, Δy is small compared to Δx, so

sin(θ) ≈ Δy / sqrt(0 + Δx^2) = Δy / Δx

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