Exercises — Derivatives of sin x, cos x — proofs from first principles

Definition. The derivative is the limit of the average slope as the interval shrinks to nothing: $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}.$ Here $f(x+h)-f(x)$ is the rise and $h$ is the run, so the fraction is a slope; the limit picks the slope of the tangent line at $x$.

Ingredients. ${\lim }_{h\to 0}\frac{\sin h}{h}=1,{\lim }_{h\to 0}\frac{\cos h-1}{h}=0.$

$\sin (x+h)=\sin x\cos h+\cos x\sin h,$ $\cos (x+h)=\cos x\cos h-\sin x\sin h.$ These come from Trigonometric addition formulas. Their job in the proof is to split the moving part $h$ off from the fixed part $x$, so the $h$-limits can act on the $h$ pieces alone.

(a) $1$. (b) $0$. (c) $-\sin x$.

Idea: the ingredient wants the same angle upstairs and downstairs. Force that by multiplying by $\frac{5}{5}$: $\frac{\sin (5h)}{h}=\frac{\sin (5h)}{5h}\cdot 5.$ As $h\to 0$, the inner angle $5h\to 0$ too, so $\frac{\sin (5h)}{5h}\to 1$. Therefore ${\lim }_{h\to 0}\frac{\sin (5h)}{h}=1\cdot 5=\boxed{5}.$

Match each sine to its own angle so both become the ingredient: $\frac{\sin (3h)}{\sin (7h)}=\frac{\sin (3h)}{3h}\cdot \frac{7h}{\sin (7h)}\cdot \frac{3h}{7h}.$ As $h\to 0$: first factor $\to 1$, second factor $\to 1$ (reciprocal of the ingredient), third factor is the constant $\frac{3}{7}$. So ${\lim }_{h\to 0}\frac{\sin (3h)}{\sin (7h)}=1\cdot 1\cdot \frac{3}{7}=\boxed{\frac{3}{7}}.$

$f^{\prime }\left(\frac{\pi }{2}\right)={\lim }_{h\to 0}\frac{\sin (\frac{\pi }{2}+h)-\sin \frac{\pi }{2}}{h}.$ Since $\sin \frac{\pi }{2}=1$ and $\cos \frac{\pi }{2}=0$, the addition formula gives $\sin (\frac{\pi }{2}+h)=\sin \frac{\pi }{2}\cos h+\cos \frac{\pi }{2}\sin h=\cos h.$ So $f^{\prime }\left(\frac{\pi }{2}\right)={\lim }_{h\to 0}\frac{\cos h-1}{h}=0.$ And $\cos \frac{\pi }{2}=0$. ✔ The two agree. (Geometrically: $\sin x$ is at its flat peak at $x=\frac{\pi }{2}$, so slope $0$.)

Setup. $\frac{d}{dx}\cos x={\lim }_{h\to 0}\frac{\cos (x+h)-\cos x}{h}.$ Step 1 — split $x$ from $h$ with the addition formula (the only algebraic tool that separates a sum-angle): $={\lim }_{h\to 0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}.$ Step 2 — group the $\cos x$ terms so each ingredient can appear: $={\lim }_{h\to 0}\left[\cos x\cdot \frac{\cos h-1}{h}-\sin x\cdot \frac{\sin h}{h}\right].$ Step 3 — $\cos x$ and $\sin x$ are constants w.r.t. $h$, pull them out and substitute the ingredients ($0$ and $1$): $=\cos x\cdot 0-\sin x\cdot 1=\boxed{-\sin x}.$ Which ingredient supplies it? The surviving $-\sin x$ comes from $\lim \frac{\sin h}{h}=1$ (the $\frac{\cos h-1}{h}$ term dies to $0$). The minus sign comes from the $-$ inside the cosine addition formula.

Look at the figure below.

On the unit circle (radius $1$), mark angle $h$ at the centre.

• Inner triangle (blue): base $1$ on the horizontal, height $=\sin h$ (the vertical drop of the point). Area $=\frac{1}{2}\cdot 1\cdot \sin h=\frac{1}{2}\sin h$.
• Sector (green wedge): a slice of the disc of angle $h$. A full disc of radius $1$ has area $\pi$; the slice is the fraction $\frac{h}{2\pi }$ of it, giving $\frac{h}{2\pi }\cdot \pi =\frac{1}{2}h$. This step needs radians — the fraction $\frac{h}{2\pi }$ only works when $h$ is measured in radians.
• Outer triangle (orange): its vertical side is the tangent line to the circle at $(1,0)$, of height $\tan h$. Why $\tan$? Because $\tan h=\frac{\text{opposite}}{\text{adjacent}}=\frac{\text{vertical side}}{1}$, so the tangent segment measures exactly how tall the outer triangle rises when its base is the radius $1$. Its area $=\frac{1}{2}\cdot 1\cdot \tan h=\frac{1}{2}\tan h$.

Geometrically the inner triangle sits inside the sector, which sits inside the outer triangle, so the areas nest: $\frac{1}{2}\sin h\le \frac{1}{2}h\le \frac{1}{2}\tan h.$ Multiply by $2$ and divide by $\sin h>0$ (valid for small $h>0$): $1\le \frac{h}{\sin h}\le \frac{1}{\cos h},$ since $\frac{\tan h}{\sin h}=\frac{1}{\cos h}$. As $h\to 0$, $\cos h\to 1$, so both ends $\to 1$ and the Squeeze Theorem traps $\frac{h}{\sin h}\to 1$, hence $\frac{\sin h}{h}\to 1$.

Use the conjugate trick — multiply by $\frac{1+\cos h}{1+\cos h}$ to convert the cosine into a sine (via $1-{\cos }^{2}h={\sin }^{2}h$): $\frac{1-\cos h}{h^2}\cdot \frac{1+\cos h}{1+\cos h}=\frac{1-{\cos }^{2}h}{h^2(1+\cos h)}=\frac{{\sin }^{2}h}{h^2(1+\cos h)}.$ Now split into ingredient pieces: $={\left(\frac{\sin h}{h}\right)}^2\cdot \frac{1}{1+\cos h}.$ As $h\to 0$: ${\left(\frac{\sin h}{h}\right)}^2\to 1^2=1$, and $\frac{1}{1+\cos h}\to \frac{1}{1+1}=\frac{1}{2}$. So ${\lim }_{h\to 0}\frac{1-\cos h}{h^2}=\boxed{\frac{1}{2}}.$ Difference from ingredient 2: ingredient 2 has denominator $h$ (one power) and equals $0$; this one has $h^2$ and equals $\frac{1}{2}$. The extra power of $h$ is exactly what rescues the $\frac{1}{2}$.

$g^{\prime }(x)={\lim }_{h\to 0}\frac{\sin (2(x+h))-\sin (2x)}{h}={\lim }_{h\to 0}\frac{\sin (2x+2h)-\sin (2x)}{h}.$ Step 1 — addition formula with the moving angle being $2h$: $\sin (2x+2h)=\sin (2x)\cos (2h)+\cos (2x)\sin (2h).$ Step 2 — subtract and group: $g^{\prime }(x)={\lim }_{h\to 0}\left[\sin (2x)\cdot \frac{\cos (2h)-1}{h}+\cos (2x)\cdot \frac{\sin (2h)}{h}\right].$ Step 3 — repair each limit to its ingredient form (angle $2h$ must match top and bottom): $\frac{\cos (2h)-1}{h}=\frac{\cos (2h)-1}{2h}\cdot 2\to 0\cdot 2=0,\frac{\sin (2h)}{h}=\frac{\sin (2h)}{2h}\cdot 2\to 1\cdot 2=2.$ Step 4 — substitute: $g^{\prime }(x)=\sin (2x)\cdot 0+\cos (2x)\cdot 2=\boxed{2\cos (2x)}.$ Where the $2$ is born: in Step 3, from $\frac{\sin (2h)}{h}\to 2$ — the inner angle's rate. This is a hand-built preview of the Chain Rule.

Repeatedly apply $\frac{d}{dx}\sin x=\cos x$ and $\frac{d}{dx}\cos x=-\sin x$: $f^{\prime }(x)=\cos x,f^{\prime \prime }(x)=-\sin x,f^{\prime \prime \prime }(x)=-\cos x,f^{(4)}(x)=\sin x.$ Pattern: each differentiation advances the wave by $90^{\circ }=\frac{\pi }{2}$: $\sin \to \cos \to -\sin \to -\cos \to \sin ,$ a 4-step cycle. Equivalently $\frac{d}{dx}\sin x=\sin \left(x+\frac{\pi }{2}\right)$. So $f^{(n)}(x)=\sin \left(x+\frac{n\pi }{2}\right)$, and the derivatives repeat with period $4$. This ties to Taylor series of sin and cos.

$\frac{d}{dx}\tan x={\lim }_{h\to 0}\frac{\tan (x+h)-\tan x}{h}={\lim }_{h\to 0}\frac{1}{h}\left[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}\right].$ Step 1 — common denominator: $={\lim }_{h\to 0}\frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{h\cos (x+h)\cos x}.$ Step 2 — the numerator is an addition formula in reverse: $\sin (x+h)\cos x-\cos (x+h)\sin x=\sin ((x+h)-x)=\sin h.$ (This is the subtraction formula from Trigonometric addition formulas — chosen because it collapses the whole numerator to a single $\sin h$.) $={\lim }_{h\to 0}\frac{\sin h}{h\cos (x+h)\cos x}.$ Step 3 — split into an ingredient times the rest: $={\lim }_{h\to 0}\frac{\sin h}{h}\cdot \frac{1}{\cos (x+h)\cos x}=1\cdot \frac{1}{\cos x\cdot \cos x}=\frac{1}{{\cos }^{2}x}=\boxed{{\sec }^{2}x}.$ This is exactly the result in Derivatives of tan, sec, csc, cot, built here from raw ingredients.

Match each sine to its own angle: $\frac{\sin (ah)}{\sin (bh)}=\frac{\sin (ah)}{ah}\cdot \frac{bh}{\sin (bh)}\cdot \frac{ah}{bh}.$ As $h\to 0$, both inner angles $\to 0$, so $\frac{\sin (ah)}{ah}\to 1$ and $\frac{bh}{\sin (bh)}\to 1$. The last factor is the constant $\frac{a}{b}$. Hence ${\lim }_{h\to 0}\frac{\sin (ah)}{\sin (bh)}=\frac{a}{b}.$ Degenerate case $b=0$: then the denominator is $\sin (0)=0$ for every $h$, so the expression $\frac{\sin (ah)}{0}$ is undefined — the limit does not exist (division by zero). The formula $\frac{a}{b}$ also breaks ($b$ in a denominator), consistently signalling the failure.

Why it's invalid — circularity. L'Hôpital replaces $\frac{\sin h}{h}$ by $\frac{(\sin h{)}^{\prime }}{(h{)}^{\prime }}=\frac{\cos h}{1}$. But knowing $(\sin h{)}^{\prime }=\cos h$ is the derivative of sine — the very theorem whose proof depends on $\frac{\sin h}{h}\to 1$. Using it to prove its own foundation is a loop: you assumed the conclusion. Shortest valid route. The unit-circle area squeeze (Exercise 3.2): $\frac{1}{2}\sin h\le \frac{1}{2}h\le \frac{1}{2}\tan h$ gives $\cos h\le \frac{\sin h}{h}\le 1$, and since $\cos h\to 1$ the Squeeze Theorem forces $\frac{\sin h}{h}\to 1$ — with no derivative of sine used anywhere.