Chapter 6 · Application of Derivatives

Exercise 6.2 · Q9: $\frac{4\sin\theta}{2+\cos\theta}-\theta$ is increasing on $\left[0,\frac\pi2\right]$

Exercise 6.2Increasing and decreasing functions · Section 6.3

The question

Prove that $y = \dfrac{4\sin\theta}{(2 + \cos\theta)} - \theta$ is an increasing function of $\theta$ in $\left[0, \dfrac{\pi}{2}\right]$ ⁠.

Exam answer

By the quotient rule,

\begin{aligned} &\frac{d}{d\theta}\left(\frac{4\sin\theta}{2 + \cos\theta}\right) \\ &= \frac{4\cos\theta(2 + \cos\theta) + 4\sin^2\theta}{(2 + \cos\theta)^2} \\ &= \frac{8\cos\theta + 4}{(2 + \cos\theta)^2} \end{aligned}

(using $\sin^2\theta + \cos^2\theta = 1$ ⁠). So

\begin{aligned} \frac{dy}{d\theta} &= \frac{8\cos\theta + 4}{(2 + \cos\theta)^2} - 1 \\ &= \frac{4\cos\theta - \cos^2\theta}{(2 + \cos\theta)^2} \\ &= \frac{\cos\theta\,(4 - \cos\theta)}{(2 + \cos\theta)^2} \end{aligned}

For $0 < \theta < \frac{\pi}{2}$ ⁠, $\cos\theta > 0$ ⁠, $4 - \cos\theta > 0$ and $(2 + \cos\theta)^2 > 0$ ⁠.

So $\frac{dy}{d\theta} > 0$ on $\left(0, \frac{\pi}{2}\right)$ ⁠, and $y$ is continuous on $\left[0, \frac{\pi}{2}\right]$ ⁠.

Hence $y$ is an increasing function of $\theta$ in $\left[0, \dfrac{\pi}{2}\right]$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Differentiate, put everything over one denominator, and use $\sin^2\theta + \cos^2\theta = 1$ ⁠. The derivative collapses to

\frac{dy}{d\theta} = \frac{\cos\theta\,(4 - \cos\theta)}{(2 + \cos\theta)^2}.

On $\left[0, \frac{\pi}{2}\right]$ cosine lies between 0 and 1, so every factor is $\ge 0$ ⁠. A slope that is never negative means an increasing function (Theorem 1).

Given, to find, and the plan

Given: $y = \dfrac{4\sin\theta}{2 + \cos\theta} - \theta$ ⁠, with $\theta$ in radians.
To prove: $y$ is increasing on $\left[0, \frac{\pi}{2}\right]$ ⁠: $\theta_1 < \theta_2 \Rightarrow y(\theta_1) \le y(\theta_2)$ there.
Plan: (1) Check $y$ is continuous and differentiable. (2) Quotient rule. (3) Simplify $\frac{dy}{d\theta}$ to one factorised fraction. (4) Sign on $\left[0, \frac{\pi}{2}\right]$ ⁠. (5) Theorem 1 (a), and strict increase.

Basics you need

1Increasing (Definition 1) and Theorem 1

$f$ is increasing on I if $x_1 < x_2$ gives $f(x_1) \le f(x_2)$ ⁠. Theorem 1 (a): if $f$ is continuous on $[a, b]$ ⁠, differentiable on $(a, b)$ ⁠, and $f'(x) \ge 0$ on $(a, b)$ ⁠, then $f$ is increasing on $[a, b]$ ⁠. Its proof shows $f' > 0$ gives strictly increasing. Lesson: increasing and decreasing functions

2Quotient rule

$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ wherever $v \ne 0$ (Section 5.3). Here $\frac{d}{d\theta}\sin\theta = \cos\theta$ and $\frac{d}{d\theta}\cos\theta = -\sin\theta$ (Table 5.3).

3Pythagorean identity

$\sin^2\theta + \cos^2\theta = 1$ for every $\theta$ ⁠. It lets $4\cos^2\theta + 4\sin^2\theta$ become just 4.

4Cosine on $\left[0, \frac{\pi}{2}\right]$

In the first quadrant (and on its edges) $\cos\theta$ goes from $\cos 0 = 1$ down to $\cos\frac{\pi}{2} = 0$ ⁠, so $0 \le \cos\theta \le 1$ ⁠, with $\cos\theta = 0$ only at $\frac{\pi}{2}$ ⁠. Basics: signs of sin and cos

Solution

$y$ is continuous and differentiable

Since $\cos\theta \ge -1$ ⁠, the denominator satisfies $2 + \cos\theta \ge 1 > 0$ ⁠: it is never zero. So $y$ ⁠, built from $\sin\theta$ ⁠, $\cos\theta$ and $\theta$ by sums and a quotient with non-zero denominator, is differentiable (hence continuous) at every $\theta$ ⁠, in particular on $\left[0, \frac{\pi}{2}\right]$ ⁠.

Theorem 1 has two hypotheses: continuity on the closed interval and differentiability inside. A denominator that could vanish would break both, so we check it first.
Differentiate

Quotient rule with $u = 4\sin\theta$ ⁠, $v = 2 + \cos\theta$ ⁠, $u' = 4\cos\theta$ ⁠, $v' = -\sin\theta$ ⁠:
$\begin{aligned} \frac{d}{d\theta}\left(\frac{4\sin\theta}{2 + \cos\theta}\right) &= \frac{4\cos\theta(2 + \cos\theta) + 4\sin^2\theta}{(2 + \cos\theta)^2}\\ &= \frac{8\cos\theta + 4(\cos^2\theta + \sin^2\theta)}{(2 + \cos\theta)^2}\\ &= \frac{8\cos\theta + 4}{(2 + \cos\theta)^2}. \end{aligned}$
The derivative of $-\theta$ is $-1$ ⁠, so
$\frac{dy}{d\theta} = \frac{8\cos\theta + 4}{(2 + \cos\theta)^2} - 1.$
In $u'v - uv'$ ⁠, the term $-uv'$ is $-4\sin\theta\cdot(-\sin\theta) = +4\sin^2\theta$ ⁠: two minus signs make a plus. Then the Pythagorean identity tidies $4\cos^2\theta + 4\sin^2\theta$ into 4.

$\sin^2\theta + \cos^2\theta = 1$ ⁠.
One fraction, factorised
$\begin{aligned} \frac{dy}{d\theta} &= \frac{8\cos\theta + 4 - (2 + \cos\theta)^2}{(2 + \cos\theta)^2}\\ &= \frac{8\cos\theta + 4 - 4 - 4\cos\theta - \cos^2\theta}{(2 + \cos\theta)^2}\\ &= \frac{4\cos\theta - \cos^2\theta}{(2 + \cos\theta)^2}\\ &= \frac{\cos\theta\,(4 - \cos\theta)}{(2 + \cos\theta)^2}. \end{aligned}$
Writing 1 as $\frac{(2 + \cos\theta)^2}{(2 + \cos\theta)^2}$ puts everything over one denominator. After expanding $(2 + \cos\theta)^2 = 4 + 4\cos\theta + \cos^2\theta$ the constants cancel, and the factor $\cos\theta$ comes out. Now the sign can be read factor by factor.
Sign on $\left[0, \frac{\pi}{2}\right]$

For $0 \le \theta \le \frac{\pi}{2}$ we have $0 \le \cos\theta \le 1$ ⁠. Hence
- $\cos\theta \ge 0$ ⁠, and $\cos\theta > 0$ for $0 \le \theta < \frac{\pi}{2}$ ⁠;
- $4 - \cos\theta \ge 4 - 1 = 3 > 0$ ⁠;
- $(2 + \cos\theta)^2 \ge 2^2 = 4 > 0$ ⁠.
So $\frac{dy}{d\theta} \ge 0$ on $\left[0, \frac{\pi}{2}\right]$ ⁠, and in fact $\frac{dy}{d\theta} > 0$ on $\left(0, \frac{\pi}{2}\right)$ ⁠; it is 0 only at $\theta = \frac{\pi}{2}$ ⁠.

Sign chart on $\left[0, \frac{\pi}{2}\right]$ (filled dots: both ends included). The numbers at the ends are the values of each factor at $\theta = 0$ and $\theta = \frac{\pi}{2}$ ⁠; every factor is positive inside.

Only $\cos\theta$ can reach 0 on this interval, and it does so only at the right end. Inside the interval all three factors are strictly positive.

$\cos\theta$ is positive in the first quadrant and 0 at $\frac{\pi}{2}$ ⁠. Basics: signs of sin and cos
Conclude with Theorem 1

$y$ is continuous on $\left[0, \frac{\pi}{2}\right]$ ⁠, differentiable on $\left(0, \frac{\pi}{2}\right)$ ⁠, and $\frac{dy}{d\theta} > 0$ there (Step 4). By Theorem 1 (a), $y$ is increasing on $\left[0, \frac{\pi}{2}\right]$ . Because the derivative is strictly positive inside, the Mean Value step of the book's proof gives $y(\theta_1) < y(\theta_2)$ for $0 \le \theta_1 < \theta_2 \le \frac{\pi}{2}$ ⁠: $y$ is even strictly increasing there.

$y = \frac{4\sin\theta}{2 + \cos\theta} - \theta$ on $\left[0, \frac{\pi}{2}\right]$ ⁠: rising from $(0, 0)$ to $\left(\frac{\pi}{2}, 2 - \frac{\pi}{2}\right) \approx \left(\frac{\pi}{2}, 0.429\right)$ ⁠, with slope $\frac13$ at the start and slope 0 at the right end.

Theorem 1 looks only at the open interval $\left(0, \frac{\pi}{2}\right)$ for the sign of the derivative; the zero at the endpoint $\frac{\pi}{2}$ is harmless. Just past $\frac{\pi}{2}$ ⁠, $\cos\theta < 0$ would make the derivative negative, so $\frac{\pi}{2}$ is where the increase stops.

Result

$\dfrac{dy}{d\theta} = \dfrac{\cos\theta\,(4 - \cos\theta)}{(2 + \cos\theta)^2}$ ⁠, which is positive on $\left(0, \frac{\pi}{2}\right)$ and 0 at $\frac{\pi}{2}$ ⁠. By Theorem 1, $y$ is an increasing function of $\theta$ on $\left[0, \frac{\pi}{2}\right]$ (in fact strictly increasing). (The answer key gives no answer for this proof.)

Check it with numbers

$y(0) = 0$ ⁠, $y\left(\frac{\pi}{6}\right) \approx 0.174232$ ⁠, $y\left(\frac{\pi}{4}\right) \approx 0.259417$ ⁠, $y\left(\frac{\pi}{3}\right) \approx 0.338443$ ⁠, $y\left(\frac{\pi}{2}\right) = 2 - \frac{\pi}{2} \approx 0.429204$ ⁠: rising. Slope at $\frac{\pi}{3}$ ⁠: the formula gives $\frac{0.5 \times 3.5}{2.5^2} = 0.28$ ⁠, and the central difference agrees:

\frac{y\left(\frac{\pi}{3} + 0.001\right) - y\left(\frac{\pi}{3} - 0.001\right)}{0.002} \approx 0.280000.

Common mistakes

Sign slip in the quotient rule: $-uv' = -4\sin\theta(-\sin\theta) = +4\sin^2\theta$ ⁠. Writing $-4\sin^2\theta$ blocks the Pythagorean simplification.
Forgetting the $-\theta$ term, whose derivative $-1$ is what produces the neat factorisation.
Worrying that $\frac{dy}{d\theta} = 0$ at $\frac{\pi}{2}$ ⁠. Theorem 1 tests the sign only inside the interval; an endpoint zero does not matter.

What to take away

After differentiating a trigonometric quotient, look for $\sin^2\theta + \cos^2\theta = 1$ ⁠.
Bring the derivative to one factorised fraction, then bound each factor using the interval (here $0 \le \cos\theta \le 1$ ⁠).
A zero of the derivative at an endpoint does not affect increasing on the closed interval.

Every fact used, and where it comes from

Fact	Used in	Source
Increasing and strictly increasing	Step 5	Definition 1 (i), (iv), Section 6.3 (Lesson: increasing and decreasing functions)
$\frac{d}{d\theta}\sin\theta = \cos\theta$ ⁠, $\frac{d}{d\theta}\cos\theta = -\sin\theta$	Step 2	Table 5.3
Quotient rule	Step 2	Algebra of derivatives, rule (3), Section 5.3
$\sin^2\theta + \cos^2\theta = 1$	Step 2	Basic trigonometric identity
$0 \le \cos\theta \le 1$ on $\left[0, \frac{\pi}{2}\right]$	Step 4	Unit-circle values of cosine (Basics: signs of sin and cos)
$f' \ge 0$ inside gives increasing on the closed interval; strict from the proof	Step 5	Theorem 1 (a) and its proof, Section 6.3

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Exercise 6.2 · Q9: 4sin⁡θ2+cos⁡θ−θ\frac{4\sin\theta}{2+\cos\theta}-\theta2+cosθ4sinθ​−θ is increasing on [0,π2]\left[0,\frac\pi2\right][0,2π​]

Basics you need

Solution

yyy is continuous and differentiable

Differentiate

One fraction, factorised

Sign on [0,π2]\left[0, \frac{\pi}{2}\right][0,2π​]

Conclude with Theorem 1

Every fact used, and where it comes from

Exercise 6.2 · Q9: $\frac{4\sin\theta}{2+\cos\theta}-\theta$ is increasing on $\left[0,\frac\pi2\right]$

$y$ is continuous and differentiable

Sign on $\left[0, \frac{\pi}{2}\right]$