Tutorials › AP Calculus › Why Trigonometric Derivatives Require Radians

Trigonometric and Exponential Derivatives · Tutorial 145 of 999

Why Trigonometric Derivatives Require Radians

See how angle units change the derivative of sine, and why AP Calculus uses radians in its standard trigonometric derivative formulas.

Intermediate 9 min read

What You'll Learn

  • Explain how radians and degrees represent the same angle with different numerical values
  • Convert a degree-valued input into radians before differentiating
  • Derive the factor of pi over 180 in the derivative of sine measured in degrees
  • Calculate and interpret the rate of change of sine for degree-valued inputs
  • Recognize and correct the mistake of omitting the angle-conversion factor

Why the Angle Unit Changes the Derivative

A trigonometric function can describe the same angle using different numerical inputs. For instance, an angle may be written as \(60^\circ\) or as \(\pi/3\) radians. The angle is the same, but the numbers \(60\) and \(\pi/3\) are not.

A derivative measures change in the function output per unit change in its input. If that input is measured in degrees rather than radians, one unit of input represents a different angle change. Consequently, the derivative formula must change too.

In AP Calculus, the standard notation \(\sin x\) treats \(x\) as a number of radians. With that convention, \(\frac{d}{dx}\sin x=\cos x\). If the input is a number of degrees, the derivative instead includes a factor of \(\pi/180\).

Key takeaway: The familiar formula \(\frac{d}{dx}\sin x=\cos x\) assumes that \(x\) is measured in radians. For an input measured in degrees, the derivative is \(\frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right)\).

Radians and Degrees as Input Scales

A radian measures an angle by comparing the length of an arc with the radius of its circle. One full turn is \(2\pi\) radians, which is also \(360^\circ\). Therefore, converting a degree measure \(x\) to radians gives

$$ x^\circ=\frac{\pi x}{180}\text{ radians}. $$

To discuss a function whose input is the numerical degree measure, define

$$ S(x)=\sin\left(\frac{\pi x}{180}\right). $$

Here \(x\) is a real number giving the angle in degrees. For example, \(S(60)=\sin(\pi/3)=\sqrt{3}/2\). This definition makes the unit conversion explicit: the sine function on the right takes its input in radians.

The derivative \(S'(x)\) measures the change in sine output per degree of input. It is not just the cosine of the angle, because increasing \(x\) by one changes the angle by \(\pi/180\) radians, not by one radian.

Deriving the Degree-Based Derivative

The sine derivative established for radian inputs is \(\frac{d}{du}\sin u=\cos u\). Use it with the converted input \(u=\pi x/180\). The chain rule gives the degree-based formula:

$$ S'(x) =\cos\left(\frac{\pi x}{180}\right)\frac{d}{dx}\left(\frac{\pi x}{180}\right) =\frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right). $$

The factor \(\pi/180\) is the rate at which the radian measure changes when the degree measure increases by one. The following difference-quotient argument shows the same factor directly.

1
Start with the derivative definition.
For a nonzero change \(h\) in the degree input, \(S'(x)\), if it exists, is the limit of \(\frac{S(x+h)-S(x)}{h}\) as \(h\) approaches zero.
2
Convert both angles to radians.
Set \(\theta=\pi x/180\) and \(\delta=\pi h/180\). Then \(\delta\) is the change in radian measure, and \(\delta/h=\pi/180\).
3
Apply the sine addition identity.
Since \(S(x+h)=\sin(\theta+\delta)=\sin\theta\cos\delta+\cos\theta\sin\delta\), subtracting \(S(x)=\sin\theta\) gives the two terms in the quotient below.
$$ \begin{aligned} \frac{S(x+h)-S(x)}{h} &=\sin\theta\frac{\cos\delta-1}{h} +\cos\theta\frac{\sin\delta}{h}\\ &=\frac{\pi}{180}\left( \sin\theta\frac{\cos\delta-1}{\delta} +\cos\theta\frac{\sin\delta}{\delta} \right). \end{aligned} $$

For radian inputs, the standard trigonometric limits are \(\lim_{\delta\to 0}\frac{\sin\delta}{\delta}=1\) and \(\lim_{\delta\to 0}\frac{\cos\delta-1}{\delta}=0\). Since \(h\to0\) implies \(\delta\to0\), the first term in parentheses tends to zero and the second tends to \(\cos\theta\). Thus

$$ S'(x)=\frac{\pi}{180}\cos\theta =\frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right). $$

This derivation identifies the source of the difference: the radian limits used in the sine derivative have the stated values for angles measured in radians. Changing to degrees changes the relationship between the input increment and the angle increment, and that relationship contributes \(\pi/180\).

Worked Examples

Worked Example: Differentiate Sine of a Degree Input

Let \(S(x)=\sin(x^\circ)\), where \(x\) is the numerical angle in degrees. Find \(S'(30)\).

First write the function using a radian input:

$$ S(x)=\sin\left(\frac{\pi x}{180}\right). $$

Differentiate by the chain rule. The derivative of the inner expression \(\pi x/180\) is \(\pi/180\):

$$ S'(x)=\frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right). $$

Now substitute \(x=30\):

$$ \begin{aligned} S'(30) &=\frac{\pi}{180}\cos\left(\frac{30\pi}{180}\right)\\ &=\frac{\pi}{180}\cos\left(\frac{\pi}{6}\right)\\ &=\frac{\pi}{180}\cdot\frac{\sqrt{3}}{2}\\ &=\frac{\pi\sqrt{3}}{360}. \end{aligned} $$

Check: \(S(x)\) is \(\sin(\pi x/180)\), so differentiating that same radian expression by the chain rule again gives \((\pi/180)\cos(\pi x/180)\), matching the formula used. The rate at \(30^\circ\) is \(\pi\sqrt{3}/360\) units of sine output per degree.

Worked Example: A Degree Angle That Depends on \(x\)

Let \(F(x)=\sin((3x)^\circ)\). Find \(F'(20)\).

The angle is \(3x\) degrees, so its radian measure is

$$ \frac{\pi(3x)}{180}=\frac{\pi x}{60}. $$

Thus \(F(x)=\sin(\pi x/60)\). Differentiating the converted expression gives

$$ F'(x)=\frac{\pi}{60}\cos\left(\frac{\pi x}{60}\right). $$

At \(x=20\),

$$ \begin{aligned} F'(20) &=\frac{\pi}{60}\cos\left(\frac{20\pi}{60}\right)\\ &=\frac{\pi}{60}\cos\left(\frac{\pi}{3}\right)\\ &=\frac{\pi}{60}\cdot\frac{1}{2}\\ &=\frac{\pi}{120}. \end{aligned} $$

Check: the radian input is \(u=\pi x/60\), whose derivative is \(\pi/60\). The chain rule therefore gives \((\pi/60)\cos u\), exactly the expression differentiated above. The answer is the rate of change per unit of \(x\); the angle itself changes at \(3\) degrees per unit of \(x\).

Worked Example: Converting a Degree-Valued Input with a Larger Scale

Let \(G(t)=\sin((180t)^\circ)\). Find \(G'(1/3)\).

Convert the angle to radians before differentiating:

$$ (180t)^\circ=\frac{\pi(180t)}{180}=\pi t\text{ radians}, \qquad G(t)=\sin(\pi t). $$

The inner expression \(\pi t\) has derivative \(\pi\), so

$$ G'(t)=\pi\cos(\pi t). $$

Substitute \(t=1/3\):

$$ \begin{aligned} G'\left(\frac{1}{3}\right) &=\pi\cos\left(\frac{\pi}{3}\right)\\ &=\pi\cdot\frac{1}{2}\\ &=\frac{\pi}{2}. \end{aligned} $$

Check: the degree angle \(180t\) changes at \(180\) degrees per unit of \(t\), equivalent to \(\pi\) radians per unit of \(t\). Differentiating \(\sin(\pi t)\) gives the same factor \(\pi\), confirming the result.

What the Standard Formulas Assume

The standard AP Calculus trigonometric derivative formulas use radian-valued inputs. In particular, \(\frac{d}{dx}\sin x=\cos x\) and \(\frac{d}{dx}\cos x=-\sin x\) are written for \(x\) in radians. The same unit convention applies when these functions are used inside other expressions.

If a variable represents a degree measure, convert that measure to radians before using those formulas. For a degree-valued input \(x\), the sine formula is

$$ \frac{d}{dx}\sin(x^\circ) =\frac{\pi}{180}\cos(x^\circ) =\frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right). $$

In the middle expression, \(\cos(x^\circ)\) means the cosine of the angle whose degree measure is \(x\); the final expression writes that angle in radians. The factor is necessary even though the output cosine describes the same angle. The derivative measures change per unit of the chosen input, and one degree is only \(\pi/180\) radians.

Common Mistake: Forgetting the Conversion Factor

A common error is to write \(\frac{d}{dx}\sin(x^\circ)=\cos(x^\circ)\), applying the radian formula directly to a degree-valued input. For example, that mistake would give

$$ \left.\frac{d}{dx}\sin(x^\circ)\right|_{x=0} \mathrel{\color{red}{\text{(incorrect)}}} 1. $$

The correction is to convert the input first:

$$ \begin{aligned} \sin(x^\circ)&=\sin\left(\frac{\pi x}{180}\right),\\ \left.\frac{d}{dx}\sin(x^\circ)\right|_{x=0} &=\left.\frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right)\right|_{x=0}\\ &=\frac{\pi}{180}\cos(0)\\ &=\frac{\pi}{180}. \end{aligned} $$

Check the scale: a one-degree input change corresponds to only \(\pi/180\) radians. At zero, the radian-input sine has slope \(1\) per radian, so its slope per degree is \(1\cdot\pi/180\), not \(1\).

Why This Matters

Differentiation describes rates, and a rate depends on the unit used for the input. A speed of \(60\) miles per hour and a speed of \(60\) kilometers per hour are different, even though the numerical values match. Likewise, the slope of sine per radian and its slope per degree are different.

Radians make the basic trigonometric derivative formulas especially simple: no conversion factor is needed in \(\frac{d}{dx}\sin x=\cos x\). This is why calculus formulas are stated using radians. Degree measures remain useful in applications and everyday contexts, but a degree-valued variable must be converted when applying the standard formulas.

Check Your Understanding

Use radians in the derivative formulas, and make the input unit explicit in each answer.

  1. Write the function whose value is the sine of an angle with degree measure \(x\), using a radian input.
  2. Find the derivative of \(\sin(x^\circ)\) with respect to \(x\).
  3. For \(H(x)=\sin((2x)^\circ)\), write \(H'(x)\).
  4. What factor is missing in the incorrect claim \(\frac{d}{dx}\sin(x^\circ)=\cos(x^\circ)\)?
  5. In one sentence, explain why the standard formula \(\frac{d}{dx}\sin x=\cos x\) assumes radians.

Test yourself: AP Calculus practice exams

Ready to check what you know? Work an AP-style practice exam, then open the full solutions.

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers all of AP Calculus AB & BC: full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon
Clinical Biostats