PROBLEMS ON INVERSE TRIGONOMETRIC FUNCTIONS

Problem 1 : 

Find the principal value of sin^-1(-1/2), (in radians and degrees). 

Solution : 

Let y  =  sin^-1(-1/2). 

Then, 

sin y  =  -1/2

The range of the principal value of sin^-1(x) is [-π/2, π/2] and hence, let us find y ∈ [-π/2, π/2]  such that 

sin y  =  -1/2

Clearly, y  =  -π/6.

Thus, the principal value of sin^-1(x) is -π/6. 

This corresponds to -30°. 

Problem 2 : 

Find the principal value of sin^-1(2), if it exists. 

Solution : 

Let y  =  sin^-1(2). 

Since the domain of y = sin^-1(x) is [-1, 1] and 2 ∉ [-1, 1], sin^-1(2) does not exist. 

Problem 3 : 

Find the principal value of

(i) sin^-1(1/√2),   (ii) sin^-1[sin(-π/3)],   (iii) sin^-1[sin(5π/6)] 

Solution : 

We know that

sin^-1 : [-1, 1] ----->  [-π/2, π/2] is given by sin^-1(x)  =  y

if and only if x  =  sin y for -1 ≤ x ≤ 1 and -π/2 ≤ y ≤ π/2.

Thus, 

(i) sin^-1(1/√2)  =  π/4

Since π/4 ∈ [-π/2, π/2] and sin (π/4) = 1/√2.

(ii) sin^-1[sin(-π/3)] = -π/3,  since -π/3 ∈ [-π/2, π/2]. 

(iii) sin^-1[sin(5π/6)] : 

=  sin^-1[sin(π - π/6)]

=  sin^-1[sin(π/6)]

=  sin^-1(1/2)

=  π/6

Since π/6 ∈ [-π/2, π/2]. 

Problem 4 : 

Find the domain of sin^-1(2 - 3x²).

Solution : 

We know that the domain of sin^-1(x) is [-1, 1]. 

This leads to -1 ≤ (2 - 3x²) ≤ 1, which implies 

-1 ≤ (2 - 3x²) ≤ 1

Subtract 2 from each term.  

-3 ≤ -3x² ≤ -1

Divide each term by -3.

1 ≥ x² ≥ 1/3

or 

1/3 ≤ x² ≤ 1/3

That is, 

1/√3 ≤ |x| ≤ 1

Then, 

x ∈ [-1, -1/√3] U [1/√3, 1]

Since a ≤ |x| ≤ b implies, x ∈ [-b, -a] U [a, b].

Problem 5 : 

Find the principal value of cos^-1(√3/2).

Solution : 

Let y  =  cos^-1(√3/2). 

Then, 

cos y  =  √3/2

The range of the principal value of y = cos^-1(x) is [0, π]. 

So, let us find y in [0, π] such that cos y = √3/2.

But, cos (π/6) = √3/2 and π/6 ∈ [0, π]. 

Therefore,

y  =  π/6

Thus, the principal value of cos^-1(√3/2) is π/6.

Problem 6 : 

Find the principal value of

(i) cos^-1(-1/√2), (ii) cos^-1[cos(-π/3)], (iii) cos^-1[cos(7π/6)] 

Solution : 

It is known that

cos^-1 : [-1, 1] ----->  [0, π] is given by cos^-1(x)  =  y

if and only if x  =  cos y for -1 ≤ x ≤ 1 and 0 ≤ y ≤ π.

Thus, 

(i) cos^-1(-1/√2)  =  3π/4,  since 3π/4 ∈ [0, π] and

cos (3π/4)  =  cos (π - π/4)  =  -cos(π/4)  =  -1/√2

(ii) cos^-1[cos(-π/3)]  =  cos^-1[cos(π/3)]  =  π/3, 

since -π/3 ∉ [0, π], but π/3 ∈ [0, π]. 

(iii) cos^-1[cos(7π/6)]  =  5π/6,

since cos(7π/6)  =  cos(π + π/6)  =  -√3/2  =  cos(5π/6) and 

5π/6 ∈ [0, π]

Problem 7 : 

Find the principal value of tan^-1(√3).

Solution : 

Let y  =  tan^-1(√3). 

Then, 

tan y  =  √3

Thus, y  =  π/3. 

Since π/3 ∈ [-π/2, π/2], the principal value of tan^-1(√3) is  π/3. 

Problem 8 : 

Find the principal value of

(i) tan^-1(-√3),   (ii) tan^-1[tan(3π/5)],   (iii) tan[tan^-1(209)] 

Solution : 

(i) tan^-1(-√3)  =  tan^-1[tan(-π/3)]  =  -π/3, 

since -π/3 ∈ (-π/2, π/2).

(ii) tan^-1[tan(3π/5)]

Let us find Ѳ ∈ (-π/2, π/2) such that tan Ѳ  =  tan(3π/5).

Since the tangent function has period π, 

tan(3π/5)  =   tan(3π/5 - π)  =  tan (-2π/5)

Therefore, 

tan^-1[tan(3π/5)]  =  tan^-1[tan(-2π/5)]  =  -2π/5, 

since -2π/5 ∈ (-π/2, π/2).

(iii) since tan^-1[tan(x)]  =  x, x ∈ R, we have 

tan^-1[tan(2019)]  =  2019

Problem 9 : 

Find the principal value of csc^-1(-1).

Solution : 

Let y  =  csc^-1(-1). 

Then, 

csc y  =  -1

Since the range of principal value branch of y = csc^-1x is

[-π/2, π/2] \ {0} and csc (-π/2)  =  -1,

we have y  =  -π/2.

Note that -π/2 ∈ [-π/2, π/2] \ {0}.

Thus the principal value of csc^-1(-1) is  -π/2. 

Problem 10 : 

Find the principal value of sec^-1(-2).

Solution : 

Let y  =  sec^-1(-2). 

Then, 

sec y  =  -2

The range of principal value branch of y = sec^-1x is

[0, π] \ {π/2}

Let us find y in [0, π] - {π/2} such that sec y = -2.

But, sec y = -2 -----> cos y = -1/2. 

Now,

cos y  =  -1/2  =  -cos(π/3)  =  cos(π - π/3)  =  cos(2π/3)

Therefore, 

y  =  2π/3

Since 2π/3 ∈ [0, π] \ {π/2}, the principal value of sec^-1(-2) is 2π/3.

Problem 11 : 

Find the value of sec^-1(-2√3/3).

Solution : 

Let Ѳ = sec^-1(-2√3/3).

Then, 

sec Ѳ  =  -2/√3 where Ѳ ∈ [0, π] \ {π/2}

Thus, cos Ѳ = -√3/2.

Now,

cos(5π/6)  =  cos(π - π/6)  =  -cos(π/6)  =  -√3/2

Hence, 

sec^-1(-2√3/3)  =  5π/6

Problem 12 : 

If cot^-1(1/7) = θ, find the value of cos θ.  

Solution : 

By definition, cot^-1(x) ∈ (0, π).

Therefore, cot^-1(1/7) = θ implies θ ∈ (0, π). 

But, cot^-1(1/7) = θ implies cot θ = 1/7 and hence tan θ = 7 and θ is acute. 

Using tan θ = 7/1, we construct a right triangle as shown below. 

Then, we have, 

cos θ  =  1/5√2

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