Trigonometric derivatives need radian measure

Try

A point moves around a unit circle. Its vertical coordinate is sin θ, with θ measured in radians. Near θ=0, a small angular movement produces almost the same numerical vertical movement. Near the top of the circle, the same angular movement changes height very little. Sketch both situations and predict where the vertical coordinate has its greatest positive rate of change.

Predict

Predict the slope of sine at 0 and at π/2. Then change the angle label from radians to degrees without moving the point. Should the rate per numerical angle unit stay the same? A degree is a much smaller angle than a radian, so the numerical input scale matters.

Experiment

Use the following exact reference values to sketch sine and its proposed derivative cosine on aligned axes. These are function and slope values, not secant measurements.

Radian angle	sin θ: vertical coordinate	cos θ: proposed slope
0	0	1
π/2	1	0
π	0	−1
3π/2	−1	0

At θ=0, compare the forward slope sin h/h for small positive and negative h using radian mode. Record the input unit with your results. Finite samples suggest a limit; the geometry and argument below explain why it is 1. Explore circle coordinates in the angle investigation and connect the horizontal coordinate to the vertical rate.

Understand

For a small positive angle h in radians, comparison of triangle, sector and tangent-triangle areas gives sin h ≤ h ≤ tan h. Dividing the appropriate positive quantities bounds sin h/h between cos h and 1. As h approaches zero these bounds approach 1. Symmetry supplies the same limit from negative h. This proof depends on sector area being h/2 for the unit circle, which uses radians.

The identity cos h−1=−2sin²(h/2) then gives (cos h−1)/h→0. Expand sin(x+h) with the angle-addition identity, subtract sin x, and divide by h. The quotient becomes sin x times (cos h−1)/h plus cos x times sin h/h. Taking limits gives (sin x)′=cos x. Applying the corresponding cosine addition identity gives (cos x)′=−sin x.

The remaining rules follow from reciprocal and quotient differentiation, developed fully in the product-and-quotient lesson: (tan x)′=sec²x, (cot x)′=−csc²x, (sec x)′=sec x tan x and (csc x)′=−csc x cot x. Each is valid only where the original expression is defined. Thus tangent and secant exclude zeros of cosine; cotangent and cosecant exclude zeros of sine. A pole is not a point where a large finite derivative can be substituted.

If the numerical input d is in degrees, the same vertical coordinate is F(d)=sin(πd/180). Its derivative with respect to d is (π/180)cos(πd/180). At zero the rate per degree is π/180, not 1. This inner factor is an instance of the chain rule taught later; changing angle units changes the rate’s denominator unit.

Apply

Complete the practice calculations and write the sine derivation yourself. For reciprocal trigonometric functions, show the two signs that arise when differentiating a reciprocal of cosine. Label angle units before checking a numerical slope. Written work should explain both the formula and the inputs where it applies.

Continue

Next compare exponential growth with logarithmic inverse rates. Product, quotient and chain rules will formalize the rule combinations previewed here. Knowing a derivative table is useful; independently choosing and justifying a rule remains a separate skill.