Why angle units affect derivatives
The derivative of \(\sin x\) is commonly written \(\cos x\), and the derivative of \(\cos x\) is commonly written \(-\sin x\). These formulas are correct when the input is measured in radians. If the input is measured in degrees, the derivatives include an additional factor.
This difference matters whenever a model uses degrees: for example, when an angle is recorded in degrees or when a periodic model specifies its phase in degrees. Differentiating such a model as if its input were already in radians gives the wrong rate of change.
Radian-mode and degree-mode inputs
The notation \(\sin x\), in the standard derivative formula, means that the numerical input \(x\) is in radians. To make the unit explicit, we will write \(\sin(\theta^\circ)\) when \(\theta\) is a numerical angle in degrees. Thus \(\sin(60^\circ)=\sin(\pi/3)=\sqrt{3}/2\).
If \(y=\sin(\theta^\circ)\), then \(\theta\) is the numerical degree measure that varies. A change of one unit in \(\theta\) is a change of one degree, not one radian. The derivative with respect to \(\theta\) must therefore account for the conversion between those units.
Deriving the degree-mode formulas
The radian-mode derivative formula is \(\frac{d}{dx}\sin x=\cos x\). We can use the definition of a derivative to see precisely how the formula changes when the input is a degree measure.
Let \(y=\sin(\theta^\circ)\), and let \(h\) be a change in the numerical degree measure. The angle-addition identity gives the following difference quotient:
Set \(u=\frac{\pi h}{180}\), so that \(h^\circ\) is \(u\) radians and \(h=\frac{180u}{\pi}\). The standard radian limits are \(\lim_{u\to 0}\frac{\sin u}{u}=1\) and \(\lim_{u\to 0}\frac{\cos u-1}{u}=0\). The second follows, for example, from \(\cos u-1=-2\sin^2(u/2)\). Therefore,
Taking the limit of the difference quotient now gives the derivative. The same conversion, together with the radian-mode cosine derivative, gives the corresponding formula for cosine.
The factor is not an optional adjustment or an approximation. It is the exact conversion factor between a change measured in degrees and the corresponding change measured in radians. For a model whose angle changes by some number of degrees per unit of time, the degree-to-radian conversion is needed to obtain the rate of change with respect to time.
Worked Example: Differentiating a sine with a degree input
Worked Example: Differentiating a sine with a degree input
Let \(y=\sin(\theta^\circ)\). Find \(\frac{dy}{d\theta}\) at \(\theta=60\), where \(\theta\) is measured in degrees.
First use the degree-mode formula:
Substitute \(\theta=60\):
The derivative is \(\frac{\pi}{360}\) per degree of change in the numerical input \(\theta\). As a check, write the original function using radians: \(y=\sin\!\left(\frac{\pi\theta}{180}\right)\). Differentiating this converted expression gives \(\frac{\pi}{180}\cos\!\left(\frac{\pi\theta}{180}\right)\), which at \(\theta=60\) is again \(\frac{\pi}{180}\cdot\frac12=\frac{\pi}{360}\).
Applying the conversion to a model
In an applied model, the angle may be a function of another variable, such as time. The first step is still to express the degree-valued angle in radians. Then differentiate the resulting expression with respect to the model's variable. Keep track of the units: if the input variable is time in seconds, the derivative is a rate per second.
Worked Example: A cosine model with a degree-valued phase
Suppose \(f(x)=2\cos((45x+30)^\circ)\), where \(x\) is measured in seconds. Find \(f'(2)\).
Convert the entire angle to radians:
Thus the same model in radian input is \(f(x)=2\cos\!\left(\frac{\pi x}{4}+\frac{\pi}{6}\right)\). Differentiating this expression gives
At \(x=2\), the angle in radians is
Therefore,
The rate is \(-\frac{\pi\sqrt{3}}{4}\) units of \(f\) per second. To check, apply the degree-mode cosine formula to the model's angle: differentiating \(2\cos((45x+30)^\circ)\) gives \(-2\cdot\frac{\pi}{180}\cdot45\sin((45x+30)^\circ)=-\frac{\pi}{2}\sin((45x+30)^\circ)\). At \(x=2\), this is \(-\frac{\pi}{2}\sin(120^\circ)=-\frac{\pi\sqrt{3}}{4}\), as before.
Worked Example: Finding a rate in a periodic model
A quantity is modeled by \(P(t)=100+8\sin((90t+30)^\circ)\), where \(t\) is measured in seconds and \(P\) in units. Find the rate of change at \(t=\frac13\).
Convert the degree-valued angle to radians:
The model becomes \(P(t)=100+8\sin\!\left(\frac{\pi t}{2}+\frac{\pi}{6}\right)\). Differentiating gives
At \(t=\frac13\), the angle is
It follows that
The rate is \(2\pi\) units per second. As a check, the degree-mode formula gives \(8\cdot\frac{\pi}{180}\cdot90\cos((90t+30)^\circ)=4\pi\cos((90t+30)^\circ)\). At \(t=\frac13\), the angle is \(60^\circ\), so the rate is \(4\pi\cos(60^\circ)=2\pi\) units per second.
Common mistake: using the radian formula directly
A frequent error is to see a sine in a model and immediately write its derivative as the cosine of the same displayed angle, without checking the angle's units. For example, suppose \(y=\sin((30t)^\circ)\), with \(t\) in seconds. The incorrect calculation
treats a change of one degree as if it were a change of one radian. Convert first:
The derivative is therefore
At \(t=0\), the incorrect result is \(30\cos(0^\circ)=30\), whereas the correct rate is \(\frac{\pi}{6}\cos(0^\circ)=\frac{\pi}{6}\). The discrepancy is exactly the missing conversion factor: \(30\) degrees per second is \(\frac{\pi}{6}\) radians per second.
Why the distinction matters
A derivative measures how much an output changes for a small change in its input. The numerical size of an angle change depends on the unit used to measure it: one radian is approximately \(57.3\) degrees. Consequently, a one-unit change in a degree measure is much smaller than a one-unit change in a radian measure, and the corresponding derivative has a different scale.
The radian formulas are especially convenient because their conversion factor is \(1\), which is why they appear in the standard derivative rules. Degree-based inputs are also valid, but they must be converted consistently. This is not merely a matter of notation: omitting the factor changes calculated slopes and rates.
Check Your Understanding
In each question, pay attention to whether the angle is measured in degrees or radians.
- Differentiate \(y=\cos(\theta^\circ)\) with respect to the numerical degree measure \(\theta\).
- Rewrite \(\sin((120t)^\circ)\) as a sine with radian input.
- For \(g(x)=3\sin((60x)^\circ)\), find \(g'(x)\).
- Explain why the derivative of \(\sin(\theta^\circ)\) is not simply \(\cos(\theta^\circ)\).
- A student differentiates \(q(t)=\cos((20t)^\circ)\) as \(q'(t)=-20\sin((20t)^\circ)\). Identify the error and write the corrected derivative.
