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Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q21: Continuity of sin⁡x±cos⁡x\sin x\pm\cos x and sin⁡xcos⁡x\sin x\cos x

Exercise 5.1Algebra of continuous functions · Section 5.2.1
The question

21. Discuss the continuity of the following functions:

(a) f(x)=sin⁡x+cos⁡xf(x)=\sin x+\cos x    (b) f(x)=sin⁡x−cos⁡xf(x)=\sin x-\cos x

(c) f(x)=sin⁡x . cos⁡xf(x)=\sin x\,.\,\cos x

Exam answer

Let cc be any real number and put x=c+hx = c + h⁠. Then x→cx \to c means h→0h \to 0⁠.

lim⁡x→csin⁡x=lim⁡h→0sin⁡(c+h)=lim⁡h→0(sin⁡ccos⁡h+cos⁡csin⁡h)=sin⁡c⋅1+cos⁡c⋅0=sin⁡c\begin{aligned} \lim_{x\to c}\sin x &= \lim_{h\to0}\sin(c + h) \\ &= \lim_{h\to0}\big(\sin c\cos h \\ &\qquad + \cos c\sin h\big) \\ &= \sin c\cdot1 + \cos c\cdot0 \\ &= \sin c \end{aligned} lim⁡x→ccos⁡x=lim⁡h→0cos⁡(c+h)=lim⁡h→0(cos⁡ccos⁡h−sin⁡csin⁡h)=cos⁡c⋅1−sin⁡c⋅0=cos⁡c\begin{aligned} \lim_{x\to c}\cos x &= \lim_{h\to0}\cos(c + h) \\ &= \lim_{h\to0}\big(\cos c\cos h \\ &\qquad - \sin c\sin h\big) \\ &= \cos c\cdot1 - \sin c\cdot0 \\ &= \cos c \end{aligned}

So sin⁡x\sin x and cos⁡x\cos x are continuous at every real number.

The sum, difference and product of two continuous functions are continuous.

∴\therefore (a) sin⁡x+cos⁡x\sin x + \cos x⁠, (b) sin⁡x−cos⁡x\sin x - \cos x and (c) sin⁡xcos⁡x\sin x\cos x are all continuous at every x∈Rx \in \R⁠.

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

The one idea this problem is built on

Once we know that sine and cosine are each continuous at every real number, Theorem 1 does the rest: a sum, a difference and a product of two functions continuous at a point are continuous at that point. The only real work is making sure we have proved continuity of cosine, which the textbook leaves as "a similar proof" after Example 17.

Given, to find, and the plan
Given
Three functions built from sin⁡x\sin x and cos⁡x\cos x (in (c), ".." means multiplication).
To find
The points of R\R at which each function is continuous, and any points of discontinuity.
Plan
(1) Domains. (2) Recall why sine is continuous. (3) Prove cosine is continuous. (4)–(6) Apply Theorem 1 (1), (2), (3). (7) Conclude, and cross-check by rewriting each function as a single sine.

Basics you need

1Continuous function

A function is continuous at cc when lim⁡x→cf(x)=f(c)\lim_{x\to c}f(x)=f(c)⁠; it is a continuous function when it is continuous at every point of its domain (Definitions 1 and 2, Section 5.2). Basics: what a limit is

2Theorem 1

If ff and gg are continuous at cc⁠, so are f+gf+g⁠, f−gf-g and f⋅gf\cdot g (and f/gf/g if g(c)≠0g(c)\ne0⁠). Basics: what a limit is

3Angle-sum formulas

sin⁡(c+h)=sin⁡ccos⁡h+cos⁡csin⁡h\sin(c+h)=\sin c\cos h+\cos c\sin h⁠, cos⁡(c+h)=cos⁡ccos⁡h−sin⁡csin⁡h\cos(c+h)=\cos c\cos h-\sin c\sin h⁠, and cos⁡h=1−2sin⁡2h2\cos h=1-2\sin^2\frac{h}{2}⁠. Basics: trigonometric identities

4Two basic limits

lim⁡h→0sin⁡h=0\lim_{h\to0}\sin h=0 (stated in Example 17), and from it lim⁡h→0cos⁡h=1\lim_{h\to0}\cos h=1 (Step 2). Basics: what a limit is

Solution

  1. The domains

    sin⁡x\sin x and cos⁡x\cos x are defined for every real xx⁠, so their sum, difference and product are too. All three functions have domain R\R⁠, and we must examine continuity at every real number cc⁠.

    There is no division here, so no point has to be excluded. Continuity is then a question about each c∈Rc\in\R⁠.

  2. Sine is continuous at every cc

    This is Example 17. With x=c+hx=c+h⁠:

    lim⁡x→csin⁡x=lim⁡h→0(sin⁡ccos⁡h+cos⁡csin⁡h)=sin⁡c⋅1+cos⁡c⋅0=sin⁡c.\begin{aligned} \lim_{x\to c}\sin x&=\lim_{h\to0}\left(\sin c\cos h+\cos c\sin h\right)\\ &=\sin c\cdot1+\cos c\cdot0=\sin c. \end{aligned}

    Here lim⁡h→0sin⁡h=0\lim_{h\to0}\sin h=0 is the fact stated in Example 17, and lim⁡h→0cos⁡h=1\lim_{h\to0}\cos h=1 follows from it:

    cos⁡h=1−2sin⁡2h2 ⟶ 1−2⋅02=1.\cos h=1-2\sin^2\tfrac{h}{2}\ \longrightarrow\ 1-2\cdot0^2=1.
    The example uses cos⁡h→1\cos h\to1 silently in its last line. The half-angle identity shows it follows from sin⁡→0\sin\to0 (applied at h2\frac h2⁠, which also tends to 00⁠), so nothing new is being assumed.
    cos⁡2θ=1−2sin⁡2θ\cos 2\theta=1-2\sin^2\theta with θ=h2\theta=\frac h2⁠. Basics: trigonometric identities · Example 17

  3. Cosine is continuous at every cc

    The textbook's Remark after Example 17 says "a similar proof" works. Here it is. With x=c+hx=c+h⁠:

    lim⁡x→ccos⁡x=lim⁡h→0cos⁡(c+h)=lim⁡h→0(cos⁡ccos⁡h−sin⁡csin⁡h)=cos⁡c⋅1−sin⁡c⋅0=cos⁡c.\begin{aligned} \lim_{x\to c}\cos x&=\lim_{h\to0}\cos(c+h)\\ &=\lim_{h\to0}\left(\cos c\cos h-\sin c\sin h\right)\\ &=\cos c\cdot1-\sin c\cdot0=\cos c. \end{aligned}

    So lim⁡x→ccos⁡x=cos⁡c\lim_{x\to c}\cos x=\cos c for every real cc⁠: cosine is continuous everywhere.

    cos⁡c\cos c and sin⁡c\sin c are fixed numbers once cc is fixed; only hh varies. So the limit of each term is that fixed number times the limit of cos⁡h\cos h or sin⁡h\sin h⁠, which we know from Step 2.
    cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B)=\cos A\cos B-\sin A\sin B⁠. Basics: trigonometric identities

  4. (a) sin⁡x+cos⁡x\sin x+\cos x

    Both sin⁡\sin and cos⁡\cos are continuous at every cc (Steps 2, 3). By Theorem 1 (1), their sum is continuous at every cc⁠:

    lim⁡x→c(sin⁡x+cos⁡x)=sin⁡c+cos⁡c.\lim_{x\to c}(\sin x+\cos x)=\sin c+\cos c.
    Theorem 1 (1) is exactly "limit of a sum = sum of the limits" combined with continuity of each part.
    Theorem 1 (1). Lesson: algebra of continuous functions

  5. (b) sin⁡x−cos⁡x\sin x-\cos x

    By Theorem 1 (2), the difference of two functions continuous at cc is continuous at cc⁠:

    lim⁡x→c(sin⁡x−cos⁡x)=sin⁡c−cos⁡c.\lim_{x\to c}(\sin x-\cos x)=\sin c-\cos c.
    Equivalently, −cos⁡x-\cos x is continuous (Remark (i) after Theorem 1, with λ=−1\lambda=-1⁠), and sin⁡x−cos⁡x=sin⁡x+(−cos⁡x)\sin x-\cos x=\sin x+(-\cos x) is a sum.
    Theorem 1 (2). Lesson: algebra of continuous functions

  6. (c) sin⁡x⋅cos⁡x\sin x\cdot\cos x

    By Theorem 1 (3), the product of two functions continuous at cc is continuous at cc⁠:

    lim⁡x→c(sin⁡xcos⁡x)=sin⁡ccos⁡c.\lim_{x\to c}(\sin x\cos x)=\sin c\cos c.
    The dot in "sin⁡x . cos⁡x\sin x\,.\,\cos x" is the textbook's multiplication sign; it is not a composition. A product of continuous functions is continuous.
    Theorem 1 (3). Lesson: algebra of continuous functions

  7. Conclusion, and a cross-check by a second route

    Each of the three functions is continuous at every real number cc⁠, which is its whole domain. So (a), (b) and (c) are all continuous functions, with no point of discontinuity.

    Second route: by the angle-sum formulas,

    sin⁡x+cos⁡x=2 sin⁡(x+π4),sin⁡x−cos⁡x=2 sin⁡(x−π4),sin⁡xcos⁡x=12sin⁡2x.\begin{aligned} \sin x+\cos x&=\sqrt2\,\sin\left(x+\tfrac{\pi}{4}\right),\\ \sin x-\cos x&=\sqrt2\,\sin\left(x-\tfrac{\pi}{4}\right),\\ \sin x\cos x&=\tfrac12\sin 2x. \end{aligned}

    Each right side is a constant times sine of a polynomial, continuous by Theorem 2 (composite) and Remark (i) (constant multiple). The two routes agree.

    For example, since cos⁡π4=sin⁡π4=12\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{1}{\sqrt2}⁠, the angle-sum formula gives 2sin⁡(x+π4)\sqrt2\sin\left(x+\frac{\pi}{4}\right) =sin⁡x+cos⁡x=\sin x+\cos x⁠. Seeing the same conclusion from a different theorem is a good check that nothing was missed.
    Theorem 2: a composite of continuous functions is continuous. Lesson: algebra of continuous functions
Result

sin⁡x+cos⁡x\sin x+\cos x⁠, sin⁡x−cos⁡x\sin x-\cos x and sin⁡xcos⁡x\sin x\cos x are each continuous at every real number. All three are continuous functions on R\R⁠.

(Textbook answer: (a), (b) and (c) are all continuous.)

Check it with numbers

At c=1c=1 (radians), compare the value at 11 with values at 0.9990.999 and 1.0011.001⁠:

Functionat 0.9990.999at 11at 1.0011.001
sin⁡x+cos⁡x\sin x+\cos x1.38207381.38177331.3814714
sin⁡x−cos⁡x\sin x-\cos x0.29978680.30116870.3025503
sin⁡xcos⁡x\sin x\cos x0.45506400.45464870.4542317

The neighbours differ from the middle value by less than 0.00140.0014⁠. ✓ Also 2sin⁡(1+π4)\sqrt2\sin\left(1+\frac{\pi}{4}\right) ≈1.3817733\approx1.3817733 and 12sin⁡2≈0.4546487\frac12\sin2\approx0.4546487⁠, matching the middle column. ✓

Common mistakes
  1. Assuming cosine is continuous without saying why. The textbook proves it only for sine; give the "similar proof" (Step 3) or at least cite the Remark after Example 17.
  2. Reading sin⁡x . cos⁡x\sin x\,.\,\cos x as sin⁡(cos⁡x)\sin(\cos x)⁠. The dot means multiplication. (sin⁡(cos⁡x)\sin(\cos x) is also continuous, but by Theorem 2, not Theorem 1.)
  3. Checking a few points only. "Discuss the continuity" means at every real number. Use an arbitrary cc⁠.
What to take away
  1. Sine and cosine are continuous at every real number; the key limits are sin⁡h→0\sin h\to0 and cos⁡h→1\cos h\to1⁠.
  2. Sums, differences and products of continuous functions are continuous (Theorem 1), so any expression built this way from sine, cosine and polynomials is continuous on R\R⁠.
  3. Only division can create trouble, at the zeros of the denominator: that is the subject of Q22.

Every fact used, and where it comes from

FactUsed inSource
Continuity at a point; continuous functionSteps 1, 7Definitions 1 and 2, Section 5.2 (Lesson: continuity at a point)
Sine is continuous; lim⁡h→0sin⁡h=0\lim_{h\to0}\sin h=0Step 2Example 17, Section 5.2.1 (the limit is stated there without proof)
Cosine is continuousStep 3Remark after Example 17 (proved in Step 3)
Sum, difference, product of continuous functionsSteps 4–6Theorem 1 (1)–(3), Section 5.2.1 (Lesson: algebra of continuous functions)
Constant multiple; composite of continuous functionsStep 7Remark (i) after Theorem 1; Theorem 2, Section 5.2.1
Angle-sum and half-angle formulasSteps 2, 3, 7Class XI, Chapter 3 (Trigonometric Functions)
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