Trigonometry in radians · Calculus I: Limits & Derivatives

Calculus refuses to use degrees. Not out of snobbery — the formulas simply break. \frac{d}{dx}\sin x = \cos x is true in radians and false in degrees, where the answer picks up a stray factor of \pi/180 and drags it through every subsequent line.

This lesson explains where that factor comes from, and stocks the identities you'll actually reach for.

What a radian is

A degree is arbitrary — 360 of them per circle, inherited from Babylonian astronomy and the convenience of a number with lots of divisors.

A radian is not arbitrary. One radian is the angle that subtends an arc equal in length to the radius. Since the full circumference is 2\pi r, a full turn is 2\pi radians:

$180° = \pi \text{ rad}, \qquad 1 \text{ rad} = \frac{180°}{\pi} \approx 57.3°$

The payoff, from §0.1:

$\text{arc length} = r\theta, \qquad \text{sector area} = \tfrac12 r^2\theta$

In degrees those become \frac{\pi r\theta}{180} and \frac{\pi r^2\theta}{360}. Radians are the unit in which the conversion constant is 1 — the same reason physicists like natural units.

The unit circle

Forget right triangles for a moment; they only handle angles up to 90°. The definition that covers everything: start at (1,0) on the unit circle, walk counterclockwise an arc length of \theta, and land at

$(\cos\theta,\ \sin\theta)$

Cosine is the x-coordinate, sine is the y-coordinate. That single picture gives you:

Range. Both live in [-1, 1], because that's how wide the circle is.
Periodicity. \theta and \theta + 2\pi are the same point, so both functions repeat with period 2\pi.
Parity. Walking -\theta mirrors across the x-axis: x is unchanged and y flips, so \cos is even and \sin is odd.
The Pythagorean identity. The point is on the unit circle, so \cos^2\theta + \sin^2\theta = 1. It's the circle's equation, renamed.

Values worth knowing cold:

\theta	0	\pi/6	\pi/4	\pi/3	\pi/2	\pi
\sin	0	1/2	\sqrt2/2	\sqrt3/2	1	0
\cos	1	\sqrt3/2	\sqrt2/2	1/2	0	-1

The identities you'll actually use

Pythagorean, and its two derived forms (divide by \cos^2 and \sin^2):

$\sin^2\theta + \cos^2\theta = 1, \qquad 1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \csc^2\theta$

The second one is the engine of trigonometric substitution in §4.

Angle sum, the source of nearly everything else:

$\sin(A+B) = \sin A\cos B + \cos A \sin B$ $\cos(A+B) = \cos A\cos B - \sin A\sin B$

Double angle, which is angle-sum with A = B:

$\sin 2\theta = 2\sin\theta\cos\theta, \qquad \cos2\theta = \cos^2\theta - \sin^2\theta = 1 - 2\sin^2\theta = 2\cos^2\theta - 1$

Half angle — the last two rearranged, and the ones that make \int \sin^2 and \int\cos^2 possible at all:

$\sin^2\theta = \frac{1 - \cos2\theta}{2}, \qquad \cos^2\theta = \frac{1 + \cos2\theta}{2}$

Memorise the angle-sum pair and the Pythagorean identity. Everything in this section is five lines of algebra away from them, and re-deriving beats misremembering.

Small angles: the reason radians win

Here's the fact the next module is built on. Draw a tiny angle \theta on the unit circle and look at three lengths:

the arc from (1,0) to the point: length exactly \theta;
the vertical drop from the point to the x-axis: length \sin\theta;
the tangent segment out to where the ray meets the vertical line x=1: length \tan\theta.

The straight drop is shorter than the curved arc, and the arc is shorter than the tangent route:

$\sin\theta < \theta < \tan\theta \qquad (0 < \theta < \pi/2)$

Divide through by \sin\theta and flip:

$\cos\theta < \frac{\sin\theta}{\theta} < 1$

As \theta \to 0, \cos\theta \to 1, and \frac{\sin\theta}{\theta} is trapped between two things heading to 1. That is the squeeze theorem, and it proves

$\lim_{\theta\to0}\frac{\sin\theta}{\theta} = 1$

which §1 does properly and which is the reason \frac{d}{dx}\sin x = \cos x.

Notice the argument used arc length = \theta. In degrees the arc is \frac{\pi\theta}{180}, the sandwich reads \sin\theta^\circ < \frac{\pi\theta}{180} < \tan\theta^\circ, and the limit becomes \pi/180 instead of 1. Every derivative downstream inherits that constant. That's the whole story: radians are the unit that makes arc length equal to angle, and calculus cares about arc length.

Doing it in Python

The sandwich, tightening:

from math import sin, tan, cos

print(f"{'theta':>10} {'sin':>12} {'theta':>12} {'tan':>12} {'sin/theta':>12}")
for t in (1.0, 0.5, 0.2, 0.05, 0.01):
    print(f"{t:>10} {sin(t):>12.8f} {t:>12.8f} {tan(t):>12.8f} {sin(t)/t:>12.8f}")

print("\nsin < theta < tan at every row, and the gaps close")

The degrees trap, priced exactly:

from math import sin, radians, pi

h = 1e-7

rad_slope = (sin(0 + h) - sin(0)) / h
deg_slope = (sin(radians(0 + h)) - sin(radians(0))) / h

print(f"d/dx sin(x) at 0, x in radians : {rad_slope:.8f}")
print(f"d/dx sin(x) at 0, x in degrees : {deg_slope:.8f}")
print(f"pi/180                         : {pi/180:.8f}")
print("\nevery derivative of every trig function inherits that factor. use radians.")

Identities are checkable, and checking beats remembering:

from math import sin, cos, tan, pi

xs = (0.3, 1.1, 2.4, 4.7)
print(f"{'x':>6} {'sin2x':>12} {'2 sin cos':>12} {'cos2x':>12} {'1-2sin^2':>12}")
for x in xs:
    print(f"{x:>6} {sin(2*x):>12.8f} {2*sin(x)*cos(x):>12.8f} "
          f"{cos(2*x):>12.8f} {1 - 2*sin(x)**2:>12.8f}")

print("\nboth double-angle forms hold to machine precision")

Worked example

A Ferris wheel of radius 20 m has its centre 22 m above the ground and completes one rotation every 40 seconds. You board at the lowest point. Write your height as a function of time.

Height oscillates between 22 - 20 = 2 m and 22 + 20 = 42 m, so amplitude 20 about a midline of 22.

Period 40 s means the argument must advance by 2\pi when t advances by 40, so the coefficient of t is \frac{2\pi}{40} = \frac{\pi}{20}.

At t = 0 we're at the bottom, the minimum. -\cos starts at its minimum, so:

$h(t) = 22 - 20\cos\!\left(\frac{\pi t}{20}\right)$

Check the quarter points: h(0) = 22 - 20 = 2 ✓ (bottom). h(20) = 22 + 20 = 42 ✓ (top, half a rotation later). h(10) = 22 - 20\cos(\pi/2) = 22 ✓ (level with the centre a quarter turn in).

This is the standard A\cos(Bt + C) + D shape: A amplitude, D midline, period 2\pi/B, and C the phase shift — the transformations from §0.2 applied to a wave. In §2 differentiating it gives your vertical velocity, and the chain rule is where that \pi/20 reappears.

Your turn

1. Convert 150° to radians and \frac{5\pi}{4} to degrees.

2. A circle has radius 8 cm. Find the arc length and sector area for a central angle of \frac{3\pi}{4}.

3. Simplify \dfrac{\sin 2\theta}{1 + \cos 2\theta}.

Solutions

1. Multiply by \frac{\pi}{180} one way, \frac{180}{\pi} the other:

$150° = 150 \cdot \frac{\pi}{180} = \boxed{\frac{5\pi}{6}}, \qquad \frac{5\pi}{4} = \frac{5\pi}{4}\cdot\frac{180}{\pi} = \boxed{225°}$

2. Both formulas need radians, and \frac{3\pi}{4} already is:

$s = r\theta = 8 \cdot \frac{3\pi}{4} = 6\pi \approx 18.85 \text{ cm}$

$A = \tfrac12 r^2\theta = \tfrac12 \cdot 64 \cdot \frac{3\pi}{4} = 24\pi \approx 75.40 \text{ cm}^2$

Sanity check: \frac{3\pi}{4} is three-eighths of a full turn. Three-eighths of the circumference 16\pi is 6\pi ✓, and of the area 64\pi is 24\pi ✓.

3. Use the double-angle forms, choosing the version of \cos2\theta that makes the denominator collapse — \cos2\theta = 2\cos^2\theta - 1:

$\frac{2\sin\theta\cos\theta}{1 + 2\cos^2\theta - 1} = \frac{2\sin\theta\cos\theta}{2\cos^2\theta} = \frac{\sin\theta}{\cos\theta} = \boxed{\tan\theta}$

Picking the right form of \cos 2\theta out of the three is the entire trick, and the rule is simple: pick the one that cancels whatever else is in the expression.

Check yourself in code

Verify the small-angle sandwich \sin\theta < \theta < \tan\theta and watch \frac{\sin\theta}{\theta} close in on 1.

For \theta = 1, 0.1, 0.01, 0.001, print \sin\theta, \tan\theta, the ratio \frac{\sin\theta}{\theta} (all to 8 decimals), and whether the sandwich holds.

Print exactly this:

theta=1.0     sin=0.84147098  tan=1.55740772  ratio=0.84147098  sandwich=True
theta=0.1     sin=0.09983342  tan=0.10033467  ratio=0.99833417  sandwich=True
theta=0.01    sin=0.00999983  tan=0.01000033  ratio=0.99998333  sandwich=True
theta=0.001   sin=0.00100000  tan=0.00100000  ratio=0.99999983  sandwich=True

from math import sin, tan

for t in (1.0, 0.1, 0.01, 0.001):
    # print sin, tan, sin/theta, and whether sin(t) < t < tan(t)
    print(f"theta={t:<7} ...")

from math import sin, tan

for t in (1.0, 0.1, 0.01, 0.001):
    ok = sin(t) < t < tan(t)
    print(f"theta={t:<7} sin={sin(t):.8f}  tan={tan(t):.8f}  "
          f"ratio={sin(t)/t:.8f}  sandwich={ok}")

A radian is the angle whose arc equals the radius, which makes arc length r\theta and sector area \frac12 r^2\theta with no conversion constant. The unit circle defines sine and cosine beyond right triangles and hands you periodicity, parity and the Pythagorean identity for free. And the sandwich \sin\theta < \theta < \tan\theta — true only in radians — is what makes \frac{\sin\theta}{\theta} \to 1, and therefore what makes trigonometric calculus work at all.

That closes the on-ramp. Next module: what "heads toward" actually means.

5. Trigonometry in radians