3

I'm lookoing at leonbloy's answer here: Intuitive understanding of the derivatives of $\sin x$ and $\cos x$

enter image description here

Can somebody explain to me why $\phi=\theta$? Why is it that the triangles in question are similar? Thank you

Blue
  • 75,673

3 Answers3

2

Diagram

Please excuse the MS Paint usage - I have no other software at my disposal. The angle I have labelled $a$ is $\pi/2-\theta$, and so clearly $b=\theta$. Now notice the tangent line formed by the hypotenuse $h$, which is only a true tangent in the limit as $\delta\theta\to0$. Tangents are perpendicular to a circle's radius, and so the angle adjacent to $b$, inside the triangle including $\phi$, must add up to $\pi/2$ along with $b$. Therefore that angle inside the triangle with $\phi$ is $a=\pi/2-\theta$. Notice now that $\phi+a=\pi/2\therefore\phi=\theta$ as $\delta\theta\to0$.

FShrike
  • 40,125
1

@leonbloy does not say $\phi=\theta$. If you perform exact calculations, we have, equating the angle between the centre and the two points on circle: $$\phi+(\frac {\pi}{2}-\theta-\Delta \theta)=\frac {\pi}{2}-\frac {\Delta \theta}{2}$$ So obviously, for $\Delta \theta\to 0$, $\theta\to \phi$. But actually $\phi=\theta+\frac {\Delta \theta}{2}$.

Ritam_Dasgupta
  • 5,992
  • 2
  • 8
  • 23
0

Modified

Light green :- $\theta$
pink :- $\pi/2-\theta$
I am sorry if I misunderstood your question but hope the new figure helps. It's just angular properties of triangles and crossed lines.

hawexrutile
  • 65
  • 1
  • 8