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Кузбасский государственный технический университет

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1.3.

INVERSE TRIGONOMETRIC FUNCTIONS

If f(x) = cos x, prove that

cos h

− sin x

−

f(x)

= cos x

− 1

f(x + h)

sin h

10.

If f(x) = sin x, prove that

cos h

+ cos x

( +

−

f(x)

= sin x

− 1

f x

sin h

11.

If f(x) = cos x, prove that

x − t

−

x − t

f(x) − f(t)

= cos t

cos(x − t) −

sin t

sin(x − t)

12.

If f(x) = sin x, prove that

x − t

f(x) − f(t)

= sin t

cos(x

− t) −

+ cos t

sin(x − t)

13.

Prove that

1 − tan2 t

cos(2t) =

1 + tan2 t

14. Prove that if y = tan x2 , then

(a) cos x =	1	− u2	(b) sin x =	2u
	1	+ u2		1 + u2

1.3Inverse Trigonometric Functions

None of the trigonometric functions are one-to-one since they are periodic. In order to deﬁne inverses, it is customary to restrict the domains in which the functions are one-to-one as follows.

CHAPTER 1.

FUNCTIONS

1. y = sin x, −

≤ x ≤

, is one-to-one and covers the range −1 ≤ y ≤ 1.

Its inverse function is denoted

arcsin x, and we deﬁne y = arcsin x,

−

≤

x ≤ 1, if and only if, x = sin y, −

≤ y ≤

graph

2. y = cos x, 0 ≤ x ≤ π, is one-to-one and covers the range −1 ≤ y ≤ 1. Its inverse function is denoted arccos x, and we deﬁne y = arccos x, −1 ≤ x ≤ 1, if and only if, x = cos y, 0 ≤ y ≤ π.

graph

−∞

3. y = tan x,

−π

< x <

, is one-to-one and covers the range

y < ∞ Its inverse function is denoted arctan x, and we deﬁne y =

−∞

∞

arctan x,

< x <

, if and only if, x = tan y,

−π

< y <

graph

4. y = cot x, 0, x < π, is one-to-one and covers the range −∞ < y < ∞. Its inverse function is denoted arccot x, and we deﬁne y = arccot x, −∞ < x < ∞, if and only if x = cot y, 0 < y < π.

graph

1.3. INVERSE TRIGONOMETRIC FUNCTIONS

5. y = sec x, 0 ≤ x ≤

< x ≤ π is one-to-one and covers the range

−∞ < y ≤ −1 or 1 ≤ y < ∞. Its inverse function is denoted arcsec x,

arcsec x,

−∞

< x

1 or 1

≤

x <

∞

, if and only

and we deﬁne

≤ −

if, x = sec y, 0 ≤ y <

< y ≤ π.

graph

6. y = csc x, −2π ≤ x < 0 or 0 < x ≤ π2 , is one-to-one and covers the range −∞ < y ≤ −1 or 1 ≤ y < ∞. Its inverse is denoted arccsc x and

we deﬁne y = arccsc x, −∞ < x ≤ −1 or 1 ≤ x < ∞, if and only if, x = csc y, −2π ≤ y < 0 or 0 < y ≤ π2 .

Example 1.3.1 Show that each of the following equations is valid.

(a)arcsin x + arccos x = π2

(b)arctan x + arccot x = π2

(c)arcsec x + arccsc x = π2

To verify equation (a), we let arcsin x = θ.

graph

Then x = sin θ and cos that

π2 − θ = arccos x,

π2 − θ = x, as shown in the triangle. It follows

π2 = θ + arccos x = arcsin x + arccos x.

The equations in parts (b) and (c) are veriﬁed in a similar way.

CHAPTER 1. FUNCTIONS

Example 1.3.2 If θ

= arcsin x, then compute cos θ, tan θ, cot θ, sec θ and

csc θ.

If θ is −

, 0, or

, then computations are easy.

graph

Suppose that −

< x < 0 or 0 < x <

. Then, from the triangle, we get

cos θ = √

cot θ =

√

1 − x2

−

x2,

tan θ =

√1 − x2

√

sec θ =

and csc θ =

1 − x2

Example 1.3.3 Make the given substitutions to simplify the given radical

expression and compute all trigonometric functions of θ.

√ √

(a) 4 − x2, x = 2 sin θ (b) x2 − 9, x = 3 sec θ

(c)(4 + x2)3/2, x = 2 tan θ

(a)For part (a), sin θ = x2 and we use the given triangle:

graph

Then

√

cos θ =

− x2

tan θ =

cot θ =

4 − x2

√4 − x2

sec θ =

√

csc θ =

− x2

√

Furthermore, 4 − x2 = 2 cos θ and the radical sign is eliminated.

1.3. INVERSE TRIGONOMETRIC FUNCTIONS

(b)For part (b), sec θ = x3 and we use the given triangle:

graph

Then,

√

sin θ =

− 4

cos θ =

, tan θ =

x2 − 4

cot θ =

√

csc θ =

√

− 9

x2 − 9

√

Furthermore, x2 − 9 = 3 tan θ and the radical sign is eliminated.

(c)For part (c), tan θ = x2 and we use the given triangle:

graph

Then,

√

sin θ =

cos θ =

cot θ =

+ 4

x2 + 4

√

+ 4

sec θ =

csc θ =

√

Furthermore, x2 + 4 = 2 sec θ and hence

(4 + x)3/2 = (2 sec θ)3 = 8 sec3 θ.

24	CHAPTER 1. FUNCTIONS

Remark 2 The three substitutions given in Example 15 are very useful in calculus. In general, we use the following substitutions for the given radicals:

√ √

(a) a2 − x2, x = a sin θ (b) x2 − a2, x = a sec θ

√

(c)a2 + x2, x = a tan θ.

Exercises 1.3
1. Evaluate each of the following:
					√				!
1						3
(a) 3 arcsin		+ 2 arccos
	2				2
1							1
(b) 4 arctan √				+ 5arccot			√
			3							3
					2
(c) 2arcsec (−2) + 3 arccos −√
								3

(d)cos(2 arccos(x))

(e)sin(2 arccos(x))

2.Simplify each of the following expressions by eliminating the radical by using an appropriate trigonometric substitution.

(a)

(b)

3 + x

(c)

x − 2

√9 − x2

√16 + x2

x√x2 − 25

(d)

1 + x

(e)

2 − 2x

√x2 + 2x + 2

√x2 − 2x − 3

(Hint: In parts (d) and (e), complete squares ﬁrst.)

3.Some famous polynomials are the so-called Chebyshev polynomials, deﬁned by

Tn(x) = cos(n arccos x), −1 ≤ x ≤ 1, n = 0, 1, 2, . . . .

(x) + T|m−n|(x)]

1.3. INVERSE TRIGONOMETRIC FUNCTIONS

(a) Prove the recurrence relation for Chebyshev polynomials:

Tn+1(x) = 2xTn(x) − Tn−1(x) for each n ≥ 1.

(b)Show that T0(x) = 1, T1(x) = x and generate T2(x), T3(x), T4(x) and T5(x) using the recurrence relation in part (a).

(c)Determine the zeros of Tn(x) and determine where Tn(x) has its absolute maximum or minimum values, n = 1, 2, 3, 4, ?.

(Hint: Let θ = arccos x, x = cos θ. Then Tn(x) = cos(nθ), Tn+1(x) = cos(nθ + θ), Tn−1(x) = cos(nθ − θ). Use the expansion formulas and then make substitutions in part (a)).

4.Show that for all integers m and n,

Tn(x)Tm(x) = 2 [Tm+n

(Hint: use the expansion formulas as in problem 3.) 5. Find the exact value of y in each of the following

y = arcsin

√

y = arccos −2

y = arccot −

√

y = arcsec (−√

y = arcsec −√2

y = arccsc −√2

y = arccsc (−2)

y = arctan √−

√

c) y = arctan(− 3)

√

f)y = arccsc (− 2)

i)y = arcsec (−2)

√

l) y = arccot (− 3)

6. Solve the following equations for x in radians (all possible answers).

a)	2 sin4 x = sin2 x	b)	2 cos2 x − cos x − 1 = 0
c)	sin2 x + 2 sin x + 1 = 0	d)	4 sin2 x + 4 sin x + 1 = 0

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