Wemazi

Topic 2.10

Evaluate a trigonometric slope

Builds on differentiating cotangent and cosecant. 7 minutes.

Slopes at exact angles

The top graph shows \(f(x) = \tan x - 2x\), with \(x\) in radians, green where it rises and red where it falls. By the rule for tan, its slope is \(f'(x) = \sec^2 x - 2\), and the bottom graph shows it as an orange bar.

To find the slope at one input, substitute the input into this formula. The angles \(0\), \(\frac{\pi}{6}\), \(\frac{\pi}{4}\) and \(\frac{\pi}{3}\) have exact cosines, so they give exact slopes.

Go to \(x = \frac{\pi}{4}\). Here \(\cos x = \frac{\sqrt2}{2}\), so \(\sec x = \sqrt2\) and \(\sec^2 x = 2\). The slope is \(2 - 2 = 0\), and the tangent is flat.

Go to \(x = \frac{\pi}{3}\). Here \(\cos x = \frac12\), so \(\sec^2 x = 4\) and the slope is 2. At \(x = 0\), the slope is \(1 - 2 = -1\).

A slope at an exact angle

The left picture shows the unit circle, which has radius 1. Its orange horizontal leg has length \(\cos x\), and its vertical leg has length \(\sin x\). The right graph shows the slope \(\sec^2 x - 2\) from the opening.

Drag \(x\). Each of \(\tan\), \(\sec\), \(\cot\) and \(\csc\) is a ratio of the two legs.

The functions \(\tan\) and \(\sec\) divide by \(\cos x\), and \(\cot\) and \(\csc\) divide by \(\sin x\), so each has no value where its divisor is 0. The slope \(\sec^2 x - 2\) has no value at the dashed lines, where \(\cos x = 0\), so the orange dot leaves the graph as \(x\) nears them.

At every other angle, the slope is a number, which you find by substituting the exact sine and cosine.

Notation

  1. Find the derivative formula first. Then substitute the input. The input is in radians, so \(\frac{\pi}{4}\) means 45 degrees.

  2. For the standard angles, write each function with sine and cosine and use the exact values in the table.

    \[\sin\tfrac{\pi}{4} = \cos\tfrac{\pi}{4} = \tfrac{\sqrt2}{2}\]

  3. Before you substitute, test that each function in the derivative has a value at the input \(a\).

    \[\begin{aligned} \cos a &\ne 0\ (\tan,\ \sec) \\ \sin a &\ne 0\ (\cot,\ \csc) \end{aligned}\]

  4. Keep the answer exact.

    For example, \(2\sqrt3\) is exact. The decimal 3.46 is only a check.

When it fails

At \(x = \frac{\pi}{2}\), \(\cos x = 0\). The ratios \(\tan x\) and \(\sec x\) divide by 0, so they have no value there, and the red "none" in the table marks this.

The slope formula \(\sec^2 x - 2\) also divides by 0 at \(\frac{\pi}{2}\), so the slope does not exist there.

Example 1: a slope with cotangent

Find the slope of \(f(x) = 3x - \cot x\) at \(x = \frac{\pi}{4}\).

  1. At \(\frac{\pi}{4}\), \(\sin x = \frac{\sqrt2}{2}\) is not 0, so \(\cot x\) and \(\csc x\) have values there.

  2. The derivative of \(3x\) is 3, and the derivative of \(-\cot x\) is \(+\csc^2 x\).

    \[f'(x) = 3 + \csc^2 x\]

  3. At \(\frac{\pi}{4}\), \(\csc x = \frac{1}{\sqrt2/2} = \sqrt2\), so \(\csc^2 x = 2\).

    \[f'(\tfrac{\pi}{4}) = 3 + 2 = \cy{5}\]

  4. The curve is green because \(f'(x) = 3 + \csc^2 x\) is always positive, and the tangent at \(\frac{\pi}{4}\) is steep, with slope 5.

Example 2: the tangent to the secant curve

Find the tangent line to \(y = \sec x\) at \(x = \frac{\pi}{3}\).

  1. The derivative is \(\sec x\tan x\). At \(\frac{\pi}{3}\), \(\cos x = \frac12\), which is not 0, so the slope exists.

    \[y' = \sec x\tan x\]

  2. A tempting wrong move: Use \(\sec\frac{\pi}{3} = \cos\frac{\pi}{3} = \frac12\).

    \[\cr{y(\tfrac{\pi}{3}) = \tfrac12}\]

    The faint line at height 1 touches the lowest point of the curve, so \(\sec x\) is never below 1. The red dot at height \(\frac12\) lies under the curve. The secant is the reciprocal of the cosine, not the cosine.

  3. Use \(\sec\frac{\pi}{3} = \frac{1}{1/2} = 2\) and \(\tan\frac{\pi}{3} = \frac{\sqrt3/2}{1/2} = \sqrt3\). The point is \((\frac{\pi}{3}, 2)\).

    \[m = 2\cdot\sqrt3 = \cy{2\sqrt3}\]

  4. Substitute the point and the slope into the point-slope form.

    \[y - 2 = 2\sqrt3\left(x - \frac{\pi}{3}\right)\]

Practice

Find the slope of \(y = x\cot x\) at \(x = \frac{\pi}{4}\), for \(0 < x < \pi\).

The curve is red because it falls, so the slope at \(\frac{\pi}{4}\) is negative. The function is a product, so start with the product rule.

Solution

  1. Use the product rule.

    \[y' = \cot x + x\,(-\csc^2 x)\]

  2. At \(\frac{\pi}{4}\), \(\sin x \ne 0\), so the derivative has a value, with \(\cot x = 1\) and \(\csc^2 x = 2\).

    \[y'(\tfrac{\pi}{4}) = 1 - \tfrac{\pi}{4}\cdot 2 = \cy{1 - \tfrac{\pi}{2}}\]

  3. As a check, \(1 - \frac{\pi}{2} \approx -0.571\). The slope is negative, as the falling orange tangent shows.

Questions

The figure shows \(h(x) = \sec x - 2x\) between the dashed lines at \(x = \pm\frac{\pi}{2}\). Questions 1 and 2 use it. Question 3 needs a calculation.

  1. What is the slope of the tangent at the marked point \(x = \frac{\pi}{4}\)?

    • \(\sqrt2 - 2\)
    • \(2 - \sqrt2\)
    • \(\sqrt2\)
    • \(-1\)
    Answer

    \(\sqrt2 - 2\)

  2. The marked point is now at \(x = \frac{\pi}{2}\). What is \(h'(\frac{\pi}{2})\)?

    • It does not exist, because \(h(\frac{\pi}{2})\) has no value
    • \(-2\)
    • 0
    Answer

    It does not exist, because \(h(\frac{\pi}{2})\) has no value

  3. Find the slope of \(y = 2\tan x + \csc x\) at \(x = \frac{\pi}{6}\).

    • \(\tfrac83 - 2\sqrt3\)
    • \(\tfrac83 + 2\sqrt3\)
    • \(\tfrac43 - 2\sqrt3\)
    • \(\tfrac83 - \sqrt3\)
    Answer

    \(\tfrac83 - 2\sqrt3\)

Summary

To find a trigonometric slope, you find the derivative, test that it has a value at the input, and substitute the exact sine and cosine.

This works because the slope at a point is the value of the derivative there, and the standard angles have exact sines and cosines.

You use it for tangent lines, flat tangents and rates of change, with \(x\) in radians.

This is one of four skills in topic 2.10 of Which & Why: AP Calculus.