G1 - First principles for sine and cosine

G1 - First principles for sine and cosine

Derive the sine and cosine derivatives from first principles, explaining why radians matter. Use second derivatives to connect changing gradients with convexity, concavity and inflection.

The trigonometric limits in radians

Why is the gradient of y=sin⁡xy=\sin x at the origin 11 when xx is in radians, but about 0.017450.01745 when the horizontal axis counts degrees? At a radian input 0.010.01, sin⁡(0.01)/0.01≈0.999983\sin(0.01)/0.01\approx0.999983. At a degree input 0.010.01, sin⁡(0.01∘)/0.01≈0.0174533\sin(0.01^\circ)/0.01\approx0.0174533. The angle unit changes the horizontal scale and therefore the gradient.

You already know

On a unit circle, a sector of radian angle hh has area h/2h/2. A degree increment dd is πd/180\pi d/180 radians. These conversion and sector results are recall requirements, not supplied in the formula booklet.

For 0<h<π/20<h<\pi/2, compare an inscribed triangle, a sector and the outer triangle formed by the tangent to the circle. Their areas are sin⁡h/2\sin h/2, h/2h/2 and tan⁡h/2\tan h/2, so

An inscribed triangle, unit-circle sector and outer tangent triangle compare the three radian areas

sin⁡h≤h≤tan⁡h⟹cos⁡h≤sin⁡hh≤1\sin h\le h\le\tan h \quad\Longrightarrow\quad \cos h\le\frac{\sin h}{h}\le1

Both outer bounds tend to 11 as h→0+h\to0^+, forcing the middle ratio to do the same. For negative hh, both sin⁡h\sin h and hh change sign, leaving their ratio unchanged. This proves the two-sided limit without using the derivative we want to derive.

Two radian limits

lim⁡h→0sin⁡hh=1lim⁡h→0cos⁡h−1h=0\lim_{h\to0}\frac{\sin h}{h}=1 \qquad \lim_{h\to0}\frac{\cos h-1}{h}=0

The increment hh must be in radians.
Not in the formula booklet: learn it.

For the second limit, use the recall identity cos⁡h−1=−2sin⁡2(h/2)\cos h-1=-2\sin^2(h/2), which is not supplied in the formula booklet. Then, for h≠0h\ne0,

cos⁡h−1h=−sin⁡(h/2)sin⁡(h/2)h/2⟶−0×1=0\frac{\cos h-1}{h} =-\sin(h/2)\frac{\sin(h/2)}{h/2} \longrightarrow -0\times1=0

The booklet supplies the small-angle approximations sin⁡h≈h\sin h\approx h, cos⁡h≈1−h2/2\cos h\approx1-h^2/2 and tan⁡h≈h\tan h\approx h, in radians. They agree with these limits, but an approximation at a small input is different from an exact limiting statement. More generally, sin⁡(kh)/h=k[sin⁡(kh)/(kh)]→k\sin(kh)/h=k[\sin(kh)/(kh)]\to k for nonzero constant kk; for k=0k=0 the ratio is zero. This scaled limit is not printed in the booklet.

Key idea

Match the denominator to the small radian angle. With degrees, sin⁡(d∘)/d→π/180\sin(d^\circ)/d\to\pi/180, not 11; a calculator sample corroborates a limit but does not prove it.

Check both the scaling and the sign in this diagnostic question.

Deriving the sine derivative

At x=0x=0, the sine difference quotient is exactly sin⁡h/h\sin h/h, so the first limit gives gradient 11. At a general input, the addition formula separates the changing increment from the fixed angle.

Sine addition formula

sin⁡(x+h)=sin⁡xcos⁡h+cos⁡xsin⁡h\sin(x+h)=\sin x\cos h+\cos x\sin h

In the formula booklet.

Keep xx fixed throughout the limit: only hh approaches zero. Neither xx nor sin⁡x\sin x needs to be small.

Worked example 1: the full sine derivation

Question. Derive d(sin⁡x)/dxd(\sin x)/dx from first principles, with xx in radians.

Plan. Expand sin⁡(x+h)\sin(x+h), separate the two limiting ratios and use their exact limits.

f′(x)=lim⁡h→0sin⁡(x+h)−sin⁡xhfirst principles=lim⁡h→0sin⁡xcos⁡h+cos⁡xsin⁡h−sin⁡xhaddition formula=lim⁡h→0(sin⁡xcos⁡h−1h+cos⁡xsin⁡hh)group fixed factors=sin⁡x(0)+cos⁡x(1)=cos⁡xradian limits\begin{aligned} f'(x)&=\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h} && \text{first principles}\\ &=\lim_{h\to0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h} && \text{addition formula}\\ &=\lim_{h\to0}\left(\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right) && \text{group fixed factors}\\ &=\sin x(0)+\cos x(1)=\cos x && \text{radian limits} \end{aligned}

Why this step? Separating the quotient exposes the two ratios whose limits are already known, avoiding any assumption about the derivative.

Answer. d(sin⁡x)/dx=cos⁡xd(\sin x)/dx=\cos x for every real radian input.

Check. At x=π/2x=\pi/2, the gradient is 00; the sine graph has a horizontal tangent at its peak.

Change only the multiplier, from 11 to 33. Carrying it through the quotient shows why a constant multiple stays attached to the derivative.

Worked example 2: a constant multiple of sine

Question. Differentiate g(x)=3sin⁡xg(x)=3\sin x from first principles.

Plan. Factor 33 from the difference quotient and complete the same addition-formula argument.

g′(x)=lim⁡h→03sin⁡(x+h)−3sin⁡xhfirst principles=3lim⁡h→0(sin⁡xcos⁡h−1h+cos⁡xsin⁡hh)expand and group=3[sin⁡x(0)+cos⁡x(1)]apply the limits=3cos⁡xsimplify\begin{aligned} g'(x)&=\lim_{h\to0}\frac{3\sin(x+h)-3\sin x}{h} && \text{first principles}\\ &=3\lim_{h\to0}\left(\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right) && \text{expand and group}\\ &=3[\sin x(0)+\cos x(1)] && \text{apply the limits}\\ &=3\cos x && \text{simplify} \end{aligned}

Why this step? The height multiplier scales every change in height by 33, so it scales each chord and tangent gradient by 33.

Answer. g′(x)=3cos⁡xg'(x)=3\cos x.

Check. At x=0x=0, the quotient is 3sin⁡h/h→33\sin h/h\to3, agreeing with 3cos⁡03\cos0.

The sine derivative and its constant-multiple extension are not in the formula booklet: learn them. The supplied first-principles definition is a starting point, not a completed proof.

Watch out

Wrong: “sin⁡(x+h)=sin⁡x+sin⁡h\sin(x+h)=\sin x+\sin h, so the derivative is 11.” Sine does not split over addition. Right: use sin⁡xcos⁡h+cos⁡xsin⁡h\sin x\cos h+\cos x\sin h, then apply both limiting ratios.

Change the multiplier again, and show the argument rather than quoting the rule.

Deriving the cosine derivative

Cosine is falling at x=π/6x=\pi/6, so its gradient there must be negative. The minus sign enters through the subtraction in its addition formula.

Cosine addition formula

cos⁡(x+h)=cos⁡xcos⁡h−sin⁡xsin⁡h\cos(x+h)=\cos x\cos h-\sin x\sin h

In the formula booklet.

The two limits remain the same. It is the coefficient of sin⁡h/h\sin h/h that changes.

Worked example 3: the full cosine derivation

Question. Derive d(cos⁡x)/dxd(\cos x)/dx from first principles, with xx in radians.

Plan. Expand cos⁡(x+h)\cos(x+h), retain its minus sign and apply the two radian limits.

g′(x)=lim⁡h→0cos⁡(x+h)−cos⁡xhfirst principles=lim⁡h→0cos⁡xcos⁡h−sin⁡xsin⁡h−cos⁡xhaddition formula=lim⁡h→0(cos⁡xcos⁡h−1h−sin⁡xsin⁡hh)group fixed factors=cos⁡x(0)−sin⁡x(1)=−sin⁡xradian limits\begin{aligned} g'(x)&=\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h} && \text{first principles}\\ &=\lim_{h\to0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h} && \text{addition formula}\\ &=\lim_{h\to0}\left(\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}\right) && \text{group fixed factors}\\ &=\cos x(0)-\sin x(1)=-\sin x && \text{radian limits} \end{aligned}

Why this step? The minus sign belongs to the whole sin⁡xsin⁡h\sin x\sin h term and survives the limit.

Answer. d(cos⁡x)/dx=−sin⁡xd(\cos x)/dx=-\sin x for every real radian input.

Check. At x=π/6x=\pi/6, the gradient is −1/2-1/2, consistent with the falling curve.

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