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Trigonometric Derivatives · Tutorial 517 of 1495

Exact Trigonometric Derivative Values at Special Angles

Use unit-circle coordinates and the trigonometric derivative formulas to find exact derivative values, while checking carefully for zero denominators.

Intermediate 9 min read

What You'll Learn

  • Apply the six trigonometric derivative formulas at standard angles.
  • Use exact unit-circle coordinates to simplify derivative values.
  • Determine signs of trigonometric derivatives in different quadrants.
  • Identify where a trigonometric function or its derivative is undefined.
  • Check derivative values directly against their formulas.

From Unit-Circle Coordinates to Derivative Values

The derivative of a trigonometric function depends on the angle through sine and cosine. At standard unit-circle angles, those coordinates are exact, so derivative values can be expressed exactly too. The main task is to substitute the correct coordinates, simplify carefully, and distinguish a large value from a value that is not defined.

Throughout, angles are measured in radians. For an angle \(x\), the unit-circle point is \((\cos x,\sin x)\). The six derivative formulas below are used at points where the relevant function is defined.

Derivative formulas: For angles measured in radians, \(\sin x\) has derivative \(\cos x\), \(\cos x\) has derivative \(-\sin x\), \(\tan x\) has derivative \(\sec^2 x\), \(\cot x\) has derivative \(-\csc^2 x\), \(\sec x\) has derivative \(\sec x\tan x\), and \(\csc x\) has derivative \(-\csc x\cot x\), wherever the respective functions are defined.

Using \(\sec x=1/\cos x\) and \(\csc x=1/\sin x\), the last four formulas can also be written in terms of sine and cosine:

$$ \begin{aligned} (\tan x)'&=\frac{1}{\cos^2 x},& (\cot x)'&=-\frac{1}{\sin^2 x},\\ (\sec x)'&=\frac{\sin x}{\cos^2 x},& (\csc x)'&=-\frac{\cos x}{\sin^2 x}. \end{aligned} $$

These forms make the domain restrictions visible. Tangent and secant are undefined when \(\cos x=0\); cotangent and cosecant are undefined when \(\sin x=0\). A derivative cannot exist at a point where its function is not defined. At other angles, the displayed derivative formulas have nonzero denominators and can be evaluated normally.

Why the Quotient Forms Work

For example, secant is the reciprocal of cosine, so \(\sec x=1/\cos x\) wherever \(\cos x\ne0\). Applying the quotient rule to this reciprocal gives the secant derivative and explains why its denominator cannot be zero:

$$ \begin{aligned} (\sec x)' &=\left(\frac{1}{\cos x}\right)'\\ &=\frac{0\cdot\cos x-1\cdot(-\sin x)}{\cos^2 x}\\ &=\frac{\sin x}{\cos^2 x}. \end{aligned} $$

The calculation requires \(\cos x\ne0\), both for the reciprocal to be defined and for the quotient rule expression to have a nonzero denominator. The same reasoning applied to \(\csc x=1/\sin x\) gives \((\csc x)'=-\cos x/\sin^2x\), where \(\sin x\ne0\). Thus the exact-value method is not a separate differentiation rule: it evaluates a valid derivative formula at the unit-circle coordinates of the angle.

Reference Values at Common Angles

The following table gives derivative values at five common angles. “Undefined” marks an angle where the original trigonometric function is not defined, so its derivative is not defined there either. All other entries follow by substituting the standard sine and cosine coordinates into the formulas.

\(x\) \((\sin x)'\) \((\cos x)'\) \((\tan x)'\) \((\cot x)'\) \((\sec x)'\) \((\csc x)'\)
\(0\) \(1\) \(0\) \(1\) Undefined \(0\) Undefined
\(\pi/6\) \(\sqrt{3}/2\) \(-1/2\) \(4/3\) \(-4\) \(2/3\) \(-2\sqrt{3}\)
\(\pi/4\) \(\sqrt{2}/2\) \(-\sqrt{2}/2\) \(2\) \(-2\) \(\sqrt{2}\) \(-\sqrt{2}\)
\(\pi/3\) \(1/2\) \(-\sqrt{3}/2\) \(4\) \(-4/3\) \(2\sqrt{3}\) \(-2/3\)
\(\pi/2\) \(0\) \(-1\) Undefined \(-1\) Undefined \(0\)

This table does not need to be memorized as a collection of unrelated answers. For an angle in another quadrant, use its unit-circle coordinates with the same formulas. The signs of the coordinates then determine the signs of the derivative values, while squared denominators remain positive whenever they are nonzero.

A Reliable Evaluation Process

At a specified angle, it is useful to separate the calculation into two tasks: first establish the sine and cosine values, then evaluate the derivatives. This prevents a sign or domain error from being hidden inside a memorized table.

1
Find the unit-circle coordinates.
Record \(\sin x\) and \(\cos x\) exactly, including their signs.
2
Check the relevant denominators.
If a formula for tangent or secant requires \(\cos x\), or one for cotangent or cosecant requires \(\sin x\), verify that this value is nonzero.
3
Substitute and simplify.
Use the derivative formulas, retain exact radicals and fractions, and simplify only after substitution.
4
Check the result.
Substitute the same coordinates into the appropriate formula once more and confirm the sign, denominator, and simplified value.

Worked Examples

Worked Example: All Six Derivatives at \(\pi/6\)

At \(x=\pi/6\), the unit-circle coordinates give \(\sin x=1/2\) and \(\cos x=\sqrt{3}/2\). Both are nonzero, so all six functions are defined there.

For sine and cosine,

$$ (\sin x)'=\cos x=\frac{\sqrt{3}}{2}, \qquad (\cos x)'=-\sin x=-\frac{1}{2}. $$

For tangent and cotangent,

$$ (\tan x)'=\frac{1}{\cos^2 x} =\frac{1}{(\sqrt{3}/2)^2} =\frac{1}{3/4} =\frac{4}{3}, $$ $$ (\cot x)'=-\frac{1}{\sin^2 x} =-\frac{1}{(1/2)^2} =-\frac{1}{1/4} =-4. $$

For secant and cosecant,

$$ (\sec x)'=\frac{\sin x}{\cos^2 x} =\frac{1/2}{3/4} =\frac{1}{2}\cdot\frac{4}{3} =\frac{2}{3}, $$ $$ (\csc x)'=-\frac{\cos x}{\sin^2 x} =-\frac{\sqrt{3}/2}{1/4} =-\frac{\sqrt{3}}{2}\cdot4 =-2\sqrt{3}. $$

Check: inserting \(\sin x=1/2\) and \(\cos x=\sqrt{3}/2\) into the six derivative formulas gives, in order, \(\sqrt{3}/2\), \(-1/2\), \(4/3\), \(-4\), \(2/3\), and \(-2\sqrt{3}\), matching each calculation.

Worked Example: Derivative Values at \(3\pi/4\)

The angle \(3\pi/4\) lies in the second quadrant. Its reference angle is \(\pi/4\), so its coordinates are \(\sin x=\sqrt{2}/2\) and \(\cos x=-\sqrt{2}/2\). Both coordinates are nonzero.

The first two derivatives are

$$ (\sin x)'=\cos x=-\frac{\sqrt{2}}{2}, \qquad (\cos x)'=-\sin x=-\frac{\sqrt{2}}{2}. $$

For the other four,

$$ (\tan x)'=\frac{1}{(-\sqrt{2}/2)^2} =\frac{1}{1/2}=2, $$ $$ (\cot x)'=-\frac{1}{(\sqrt{2}/2)^2} =-\frac{1}{1/2}=-2, $$ $$ (\sec x)'=\frac{\sqrt{2}/2}{(-\sqrt{2}/2)^2} =\frac{\sqrt{2}/2}{1/2} =\sqrt{2}, $$ $$ (\csc x)'=-\frac{-\sqrt{2}/2}{(\sqrt{2}/2)^2} =\frac{\sqrt{2}/2}{1/2} =\sqrt{2}. $$

Check: the squared denominators are both \(1/2\), so the tangent and cotangent derivatives are \(2\) and \(-2\). The secant and cosecant numerators are respectively \(\sqrt{2}/2\) and \(\sqrt{2}/2\) after the minus sign is applied, giving \(\sqrt{2}\) for each. These agree with the substituted formulas.

Worked Example: Defined and Undefined Derivatives at \(\pi/2\)

At \(x=\pi/2\), \(\sin x=1\) and \(\cos x=0\). First evaluate the derivatives whose formulas do not divide by \(\cos x\):

$$ (\sin x)'=\cos x=0, \qquad (\cos x)'=-\sin x=-1. $$

Cotangent and cosecant are defined because \(\sin x=1\ne0\). Their derivatives are

$$ (\cot x)'=-\frac{1}{\sin^2 x} =-\frac{1}{1^2}=-1, \qquad (\csc x)'=-\frac{\cos x}{\sin^2 x} =-\frac{0}{1}=0. $$

Tangent and secant are not defined at this angle because both require division by \(\cos x=0\). Therefore \((\tan x)'\) and \((\sec x)'\) are also undefined at \(x=\pi/2\). Check: the four defined values follow directly from their formulas with nonzero denominators; the other two formulas have zero denominators, consistent with the original functions being undefined.

Common Pitfall: Treating a Zero Denominator as a Derivative Value

A common error is to substitute \(\cos(\pi/2)=0\) into \((\tan x)'=1/\cos^2x\) and report an “infinite derivative.” But the substitution produces division by zero:

$$ \frac{1}{\cos^2(\pi/2)} =\frac{1}{0^2} =\frac{1}{0}, $$

and \(1/0\) is not a real number. More fundamentally, \(\tan(\pi/2)\) itself is undefined, so there is no derivative of tangent at that point. The correct conclusion is undefined, not a finite exact value and not a value obtained by dividing by zero. Always check the function's domain before evaluating its derivative formula.

Why Exact Values Are Useful

Exact derivative values describe slopes without introducing rounding error. For example, \((\sec x)'=2/3\) at \(\pi/6\) is an exact statement, while a decimal approximation would only report a rounded version. Exact values also make signs and zero-denominator conditions transparent, which is especially helpful when comparing slopes at angles in different quadrants or using derivative values in a later calculation.

The essential pattern is simple: get the exact unit-circle coordinates, use the correct derivative formula, and verify that every denominator is nonzero. The formulas apply consistently around the circle; the angle changes the sine and cosine values, not the rules themselves.

Check Your Understanding

Use exact values throughout. For each derivative, note whether the original function is defined at the specified angle.

  1. Find \((\sin x)'\), \((\cos x)'\), and \((\tan x)'\) at \(x=\pi/3\).
  2. Evaluate \((\cot x)'\), \((\sec x)'\), and \((\csc x)'\) at \(x=5\pi/6\).
  3. Which of the six trigonometric functions have derivatives defined at \(x=0\)? Give the defined derivative values.
  4. Explain why \((\tan x)'\) is undefined at \(x=\pi/2\), rather than having a value obtained by substituting into \(1/\cos^2x\).
  5. At \(x=5\pi/4\), determine the signs of all six derivatives before calculating their exact values.

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