Tangent Slope on y = sin x
Limits and Derivatives · Class XI
Slide a point along sin x. The tangent uses slope cos(x₀) , confirm d/dx sin x = cos x.
- f(x₀)0.8415
- f′(x₀)0.5403
- y = sin x
- Tangent at x₀
Formulas in this lab
- f(x₀)
- f′(x₀)
Frequently asked questions
▶Why is the derivative of $\sin x$ equal to $\cos x$?
From first principles, $\frac{d}{dx}\sin x = \lim_{h\to 0}\frac{\sin(x+h) - \sin x}{h}$. Using $\sin(x+h) - \sin x = 2\cos(x + h/2)\sin(h/2)$ and $\lim_{h\to 0}\frac{\sin(h/2)}{h/2} = 1$, you get $\cos x$. This is the foundational trig derivative.
▶How do I use the $\sin x$ derivative lab?
Slide the point $x_0$ and the lab draws the tangent to $y = \sin x$. It reports slope $\cos(x_0)$ and the tangent line. At $x_0 = 0$, slope is $1$; at $x_0 = \pi/2$, slope is $0$ — the peak of the sine curve.
▶Derivative chain: how does $\cos x$, $-\sin x$, $-\cos x$, $\sin x$ cycle work?
$\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$. So the fourth derivative of $\sin x$ returns to $\sin x$. JEE Main loves this cycle for higher-order derivative MCQs. The lab visualizes just the first step; chain the rest mentally.
▶Where does $\cos x = \sin' x$ show up in real life?
If position is $x(t) = A\sin(\omega t)$ (SHM), velocity is $v(t) = A\omega\cos(\omega t)$ — exactly the derivative scaled by $\omega$. Maximum speed is $A\omega$, when displacement passes through zero. Music, AC circuits and pendulums all rely on this $\sin \to \cos$ derivative.