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Chapter 6 · Application of Derivatives

Example 34: tan⁡−1(sin⁡x+cos⁡x)\tan^{-1}(\sin x+\cos x) is increasing on (0,π4)\left(0,\frac{\pi}{4}\right)

ExamplesIncreasing and decreasing functions · Section 6.3 · Miscellaneous Examples
The question

Show that the function ff given by

f(x)=tan⁡−1(sin⁡x+cos⁡x),x>0f(x) = \tan^{-1}(\sin x + \cos x), \quad x > 0

is always an increasing function in (0,π4)\left(0, \frac{\pi}{4}\right)⁠.

Exam answer

By the chain rule, using (sin⁡x+cos⁡x)2=1+sin⁡2x(\sin x + \cos x)^2 = 1 + \sin 2x⁠,

f′(x)=cos⁡x−sin⁡x1+(sin⁡x+cos⁡x)2=cos⁡x−sin⁡x2+sin⁡2x\begin{aligned} f'(x) &= \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \\ &= \frac{\cos x - \sin x}{2 + \sin 2x} \end{aligned}

For 0<x<π40 < x < \frac{\pi}{4}⁠: tan⁡x<1\tan x < 1⁠, so cos⁡x>sin⁡x\cos x > \sin x and cos⁡x−sin⁡x>0\cos x - \sin x > 0⁠. Also 2+sin⁡2x>02 + \sin 2x > 0⁠.

∴ f′(x)>0\therefore\ f'(x) > 0 for all x∈(0,π4)x \in \left(0, \frac{\pi}{4}\right)⁠.

Hence ff is increasing in (0,π4)\left(0, \dfrac{\pi}{4}\right)⁠.

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

The one idea this problem is built on

To show a function is increasing on an interval, show its derivative is positive at every point of the interval (Theorem 1, Section 6.3). The chain rule gives a fraction, f′(x)=cos⁡x−sin⁡x2+sin⁡2xf'(x) = \frac{\cos x - \sin x}{2 + \sin 2x}⁠. Its denominator is positive for every xx⁠, so the sign of f′f' is the sign of cos⁡x−sin⁡x\cos x - \sin x⁠, and on (0,π4)\left(0, \frac{\pi}{4}\right) the cosine is the larger of the two.

Given, to show, and the plan
Given
f(x)=tan⁡−1(sin⁡x+cos⁡x)f(x) = \tan^{-1}(\sin x + \cos x) for x>0x > 0⁠.
To show
ff is increasing on (0,π4)\left(0, \frac{\pi}{4}\right)⁠: for any x1<x2x_1 < x_2 in that interval, f(x1)≤f(x2)f(x_1) \le f(x_2) (we will in fact get <<⁠).
Plan
Differentiate by the chain rule; simplify 1+(sin⁡x+cos⁡x)21 + (\sin x + \cos x)^2 to 2+sin⁡2x2 + \sin 2x⁠; show the denominator is positive and the numerator cos⁡x−sin⁡x\cos x - \sin x is positive on the interval; conclude with Theorem 1.

Basics you need

1Derivative of tan⁡−1\tan^{-1}

ddutan⁡−1u=11+u2\frac{d}{du}\tan^{-1}u = \frac{1}{1 + u^2} for every real uu (Chapter 5). With the chain rule, ddxtan⁡−1(u(x))=u′(x)1+u(x)2\frac{d}{dx}\tan^{-1}(u(x)) = \frac{u'(x)}{1 + u(x)^2}⁠.

2Two identities

sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 and 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x⁠. Together: (sin⁡x+cos⁡x)2=1+sin⁡2x(\sin x + \cos x)^2 = 1 + \sin 2x⁠.

3Signs and sizes of sin and cos on (0,π4)\left(0, \frac{\pi}{4}\right)

Both are positive there, and tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x} increases from 0 to 1 as xx goes from 0 to π4\frac{\pi}{4}⁠. So tan⁡x<1\tan x < 1⁠, that is sin⁡x<cos⁡x\sin x < \cos x⁠. Basics: signs of sin and cos

4Theorem 1 (Section 6.3)

If ff is continuous on [a,b][a, b]⁠, differentiable on (a,b)(a, b)⁠, and f′(x)≥0f'(x) \ge 0 for each x∈(a,b)x \in (a, b)⁠, then ff is increasing on [a,b][a, b]⁠. With f′(x)>0f'(x) > 0⁠, the Mean Value Theorem step of the proof gives strictly increasing. Lesson: increasing and decreasing

Solution

  1. Where ff is defined and differentiable

    sin⁡x+cos⁡x\sin x + \cos x is a real number for every xx⁠, and tan⁡−1\tan^{-1} is defined and differentiable on all of R\R⁠. So ff is defined, continuous and differentiable for every x>0x > 0 (and at x=0x = 0 as well).

    Theorem 1 needs continuity and differentiability on the interval we study. Unlike sin⁡−1\sin^{-1} or cos⁡−1\cos^{-1}⁠, the inverse tangent accepts every real input, so there is nothing to exclude.
    Domain of tan⁡−1\tan^{-1} is R\R⁠, range (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) (Chapter 2).

  2. Differentiate by the chain rule

    Outer function tan⁡−1u\tan^{-1}u⁠, inner function u=sin⁡x+cos⁡xu = \sin x + \cos x⁠, with u′=cos⁡x−sin⁡xu' = \cos x - \sin x⁠. So

    f′(x)=11+(sin⁡x+cos⁡x)2 (cos⁡x−sin⁡x).f'(x) = \frac{1}{1 + (\sin x + \cos x)^2}\,(\cos x - \sin x).
    This is the textbook's first line. The derivative of the outer function is evaluated at the inner function, then multiplied by the derivative of the inner function.
    ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x⁠, ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x⁠; chain rule (Chapter 5). Basics: chain rule

  3. Simplify the denominator

    (sin⁡x+cos⁡x)2=sin⁡2x+cos⁡2x+2sin⁡xcos⁡x=1+sin⁡2x,\begin{aligned} (\sin x + \cos x)^2 &= \sin^2 x + \cos^2 x + 2\sin x\cos x \\ &= 1 + \sin 2x, \end{aligned}

    so 1+(sin⁡x+cos⁡x)2=2+sin⁡2x1 + (\sin x + \cos x)^2 = 2 + \sin 2x and

    f′(x)=cos⁡x−sin⁡x2+sin⁡2x.f'(x) = \frac{\cos x - \sin x}{2 + \sin 2x}.
    This is the book's "on simplification". The simpler denominator makes its sign obvious in the next step.
    (a+b)2=a2+b2+2ab(a + b)^2 = a^2 + b^2 + 2ab⁠; sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1⁠; 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x⁠.

  4. The denominator is always positive

    For every real xx⁠, sin⁡2x≥−1\sin 2x \ge -1⁠, so

    2+sin⁡2x≥1>0.2 + \sin 2x \ge 1 > 0.

    (On (0,π4)\left(0, \frac{\pi}{4}\right) we even have 0<2x<π20 < 2x < \frac{\pi}{2}⁠, so sin⁡2x>0\sin 2x > 0 and the denominator exceeds 2.)

    A fraction with a positive denominator has the same sign as its numerator. So the whole question reduces to the sign of cos⁡x−sin⁡x\cos x - \sin x⁠. (Even without the simplification, 1+(anything)2≥11 + (\text{anything})^2 \ge 1⁠.)
    The sine of any angle lies between −1-1 and 1.

  5. The numerator is positive on (0,π4)\left(0, \frac{\pi}{4}\right)

    Let 0<x<π40 < x < \frac{\pi}{4}⁠. Then cos⁡x>0\cos x > 0⁠, and tan⁡x<tan⁡π4=1\tan x < \tan\frac{\pi}{4} = 1⁠, because tan⁡\tan increases on (0,π2)\left(0, \frac{\pi}{2}\right)⁠. Multiplying sin⁡xcos⁡x<1\frac{\sin x}{\cos x} < 1 by the positive number cos⁡x\cos x⁠:

    sin⁡x<cos⁡x,cos⁡x−sin⁡x>0.\sin x < \cos x, \qquad \cos x - \sin x > 0.
    This is the textbook's argument in different words: it writes cos⁡x>sin⁡x\cos x > \sin x as cot⁡x>1\cot x > 1⁠, which holds when tan⁡x<1\tan x < 1⁠, that is when 0<x<π40 < x < \frac{\pi}{4}⁠. Another way to see it: cos⁡x−sin⁡x=2cos⁡(x+π4)\cos x - \sin x = \sqrt{2}\cos\left(x + \frac{\pi}{4}\right)⁠, and for xx in the interval the angle x+π4x + \frac{\pi}{4} lies in (π4,π2)\left(\frac{\pi}{4}, \frac{\pi}{2}\right)⁠, where cosine is positive.
    Multiplying an inequality by a positive number keeps its direction. Basics: signs of sin and cos

  6. Conclude with Theorem 1

    By Steps 4 and 5, for every x∈(0,π4)x \in \left(0, \frac{\pi}{4}\right)⁠,

    f′(x)=cos⁡x−sin⁡x2+sin⁡2x>0.f'(x) = \frac{\cos x - \sin x}{2 + \sin 2x} > 0.

    ff is continuous on [0,π4]\left[0, \frac{\pi}{4}\right] and differentiable inside, so by Theorem 1 it is increasing there; since f′>0f' > 0 strictly, it is even strictly increasing. Restricting to the given domain x>0x > 0⁠: ff is increasing on (0,π4)\left(0, \frac{\pi}{4}\right), as required. It is also increasing on (0,π4]\left(0, \frac{\pi}{4}\right]⁠, since ff is continuous at π4\frac{\pi}{4}⁠.

    "Always increasing in (0,π4)\left(0, \frac{\pi}{4}\right)" means increasing at every point of that interval, which is what a positive derivative throughout gives. Including π4\frac{\pi}{4}⁠, where f′=0f' = 0⁠, is allowed because Theorem 1 only asks for the sign condition inside the interval.
    Theorem 1, Section 6.3. Lesson: increasing and decreasing

  7. What happens after π4\frac{\pi}{4}

    For π4<x<5π4\frac{\pi}{4} < x < \frac{5\pi}{4}⁠, the angle x+π4x + \frac{\pi}{4} lies in (π2,3π2)\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)⁠, so cos⁡x−sin⁡x<0\cos x - \sin x < 0 and f′<0f' < 0⁠: ff decreases. So (0,π4]\left(0, \frac{\pi}{4}\right] is exactly the stretch, starting from 0, on which ff increases, and x=π4x = \frac{\pi}{4} is where it is highest, f(π4)=tan⁡−12≈0.9553f\left(\frac{\pi}{4}\right) = \tan^{-1}\sqrt{2} \approx 0.9553⁠.

    π/4 π/2 3π/4 0.5 1 x y π/4 highest: tan⁻¹√2 ≈ 0.955 blue: increasing on (0, π/4) decreasing after π/4
    y=tan⁡−1(sin⁡x+cos⁡x)y = \tan^{-1}(\sin x + \cos x) for 0<x≤3π40 < x \le \frac{3\pi}{4}⁠. It rises (blue) up to x=π4x = \frac{\pi}{4}⁠, where sin⁡x+cos⁡x=2\sin x + \cos x = \sqrt{2} is largest, and falls afterwards. The open dot marks x=0x = 0⁠, which is outside the domain x>0x > 0⁠.
    This shows why the question stops at π4\frac{\pi}{4}⁠: the interval cannot be enlarged on the right.
    sin⁡π4+cos⁡π4=22+22=2\sin\frac{\pi}{4} + \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}⁠.
Result

f′(x)=cos⁡x−sin⁡x2+sin⁡2xf'(x) = \dfrac{\cos x - \sin x}{2 + \sin 2x}⁠. The denominator is always positive and cos⁡x>sin⁡x\cos x > \sin x for 0<x<π40 < x < \frac{\pi}{4}⁠, so f′(x)>0f'(x) > 0 throughout (0,π4)\left(0, \frac{\pi}{4}\right)⁠. Hence ff is increasing (indeed strictly increasing) on (0,π4)\left(0, \frac{\pi}{4}\right), and on (0,π4]\left(0, \frac{\pi}{4}\right]⁠. (This is what the textbook shows.)

Check it with numbers

Values of ff (to 4 places): f(0.1)=0.8306f(0.1) = 0.8306⁠, f(0.3)=0.8964f(0.3) = 0.8964⁠, f(0.5)=0.9357f(0.5) = 0.9357⁠, f(0.7)=0.9536f(0.7) = 0.9536⁠, f(π4)=0.9553f\left(\frac{\pi}{4}\right) = 0.9553⁠: rising. After the peak, f(1.0)=0.9443f(1.0) = 0.9443 and f(1.2)=0.9130f(1.2) = 0.9130⁠: falling. The formula gives f′(0.5)=cos⁡0.5−sin⁡0.52+sin⁡1=0.1401f'(0.5) = \frac{\cos 0.5 - \sin 0.5}{2 + \sin 1} = 0.1401⁠, and the difference quotient f(0.50001)−f(0.49999)0.00002\frac{f(0.50001) - f(0.49999)}{0.00002} is also 0.1401.

Common mistakes
  1. Forgetting the inner derivative. ddxtan⁡−1(sin⁡x+cos⁡x)\frac{d}{dx}\tan^{-1}(\sin x + \cos x) is not 11+(sin⁡x+cos⁡x)2\frac{1}{1 + (\sin x + \cos x)^2} alone; multiply by cos⁡x−sin⁡x\cos x - \sin x⁠.
  2. Sign slip in the inner derivative. ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x⁠, so the numerator is cos⁡x−sin⁡x\cos x - \sin x⁠, not cos⁡x+sin⁡x\cos x + \sin x⁠.
  3. Checking a few points only. Showing f′(0.5)>0f'(0.5) > 0 proves nothing about the whole interval; argue for an arbitrary xx in (0,π4)\left(0, \frac{\pi}{4}\right)⁠.
  4. Claiming ff is increasing for all x>0x > 0⁠. It decreases right after π4\frac{\pi}{4} (Step 7).
What to take away
  1. "Show increasing on an interval": compute f′f'⁠, prove f′>0f' > 0 for every point of the interval, cite Theorem 1.
  2. When f′f' is a fraction, check the denominator's sign separately; often it is always positive and only the numerator matters.
  3. Since tan⁡−1\tan^{-1} is itself increasing, ff increases exactly where sin⁡x+cos⁡x\sin x + \cos x does: composing with an increasing function keeps the direction.

Every fact used, and where it comes from

FactUsed inSource
ddutan⁡−1u=11+u2\frac{d}{du}\tan^{-1}u = \frac{1}{1 + u^2}⁠; chain ruleStep 2Chapter 5, derivatives of inverse trigonometric functions
Domain and range of tan⁡−1\tan^{-1}Step 1Chapter 2
sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1⁠, sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos xStep 3Trigonometric identities (Class 11)
tan⁡x<1\tan x < 1 for 0<x<π40 < x < \frac{\pi}{4}⁠; cos⁡x−sin⁡x=2cos⁡(x+π4)\cos x - \sin x = \sqrt{2}\cos\left(x + \frac{\pi}{4}\right)Steps 5, 7Trigonometric functions (Class 11)
Sign of f′f' decides increasing or decreasingSteps 6, 7Theorem 1, Section 6.3 (Lesson: increasing and decreasing)
Textbook working for this problemResultExample 34, Miscellaneous Examples, Chapter 6
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