ML Aggarwal Class 12 Maths Solutions Section A Chapter 05 Inverse Trigonometric Functions

Inverse Trigonometric Functions

 

5.1 Inverse Functions

To grasp the idea of an inverse function, examine these examples:

(1) Suppose f is a function where f(x) = 2x - 3, with domain Df = R and range Rf = R. When you rearrange to express x in terms of y, you get x = (y + 3)/2.

We can see that for every y in R, there is exactly one x in R such that y = f(x) = 2x - 3. A new function (called f inverse) emerges from the rule that connects each y in Rf to a unique x in Df.

The original function has these two key traits:

(i) No two different elements in Df map to the same element in Rf - meaning f is one-to-one.
(ii) For each y in Rf, there exists an x in Df such that f(x) = y - meaning f is onto.

(2) Take f as the function f(x) = x², where Df = R and Rf equals the set of all non-negative real numbers. When you rearrange to write x in terms of y, you get x = ± √y.

For any y > 0, we see two values of x exist. This means the equation does not define x as a function of y. The function f fails to be one-to-one, so we cannot express x as a function of y.

Looking at both examples, we conclude that only one-to-one functions can have inverse functions:

Let f be a one-to-one function with domain Df and range Rf. A function g: Rf → Df given by g(y) = x where f(x) = y is called the inverse of f, written as f⁻¹ (read as f inverse). A function f is invertible (or has an inverse) if and only if f is one-to-one.

Remarks

1. When a real function f: Df → Rf is invertible, then f⁻¹: Rf → Df satisfies f⁻¹(y) = x if and only if y = f(x) for all x in Df and y in Rf. The roles of x and y are swapped when moving from f to f⁻¹. The graph of f includes all pairs (x, y) where y = f(x) for every x in Df. The graph of f⁻¹ includes all pairs (y, x) where x = f⁻¹(y) for every y in Rf. A point (x, y) lies on the graph of f exactly when (y, x) lies on the graph of f⁻¹. Since (y, x) mirrors (x, y) across the line y = x, the graph of f⁻¹ can be obtained by reflecting the graph of f through the line y = x.

2. Often, a given function f may not be one-to-one over its whole domain. However, when we limit the domain to a smaller region, it may become one-to-one. When f is one-to-one on part of its domain, it has an inverse on that part only. When f is invertible in multiple parts of its domain, it possesses an inverse in each of those parts.

 

5.2 Inverse Trigonometric Functions

1. Inverse sine function

Think about the sine function f defined by f(x) = sin x, where Df = R and Rf = [-1, 1]. The sine function does not have the one-to-one property across its whole domain (a horizontal line at y = 1/2, for instance, intersects the curve at many points). However, when we restrict the domain to the closed interval [-π/2, π/2], the function becomes one-to-one. This means the function y = f(x) = sin x with domain [-π/2, π/2] and range [-1, 1] has an inverse function, which we call the inverse sine function or arc sine function, denoted sin⁻¹.

The relationship is: y = sin⁻¹ x if and only if x = sin y and y is in [-π/2, π/2].

The inverse sine function has these key traits:

(i) The domain of sin⁻¹ x is [-1, 1], and its range is [-π/2, π/2].
(ii) sin(sin⁻¹ x) = x for x in [-1, 1], which also means |x| ≤ 1.
(iii) sin⁻¹(sin y) = y for y in [-π/2, π/2], meaning |y| ≤ π/2.
(iv) The function sin⁻¹: [-1, 1] → [-π/2, π/2] grows steadily and is one-to-one.

Remarks

1. Besides the interval [-π/2, π/2], other intervals exist where sine is one-to-one and therefore has an inverse. However, sin⁻¹ x here will always refer to the function sin⁻¹: [-1, 1] → [-π/2, π/2] described above (unless mentioned otherwise). The portion of the curve where -π/2 ≤ y ≤ π/2 is known as the principal value branch of y = sin⁻¹ x, and these y values are called the principal values.

2. The graph of the inverse sine function comes from the graph of the original function by swapping the roles of x and y. If (a, b) is on the sine function's graph, then (b, a) is on the inverse sine function's graph. The inverse sine graph is a mirror picture of the original function along the line y = x. You can see this by comparing the graphs of y = sin x and y = sin⁻¹ x on the same axes.

3. Besides [-π/2, π/2], other regions such as [π/2, 3π/2], [3π/2, 5π/2], and [-3π/2, -π/2] exist where the sine function is one-to-one and hence has an inverse. The range of these other branches are [π/2, 3π/2], [3π/2, 5π/2], and [-3π/2, -π/2], respectively.

2. Inverse cosine function

Think about the cosine function f defined by f(x) = cos x, where Df = R and Rf = [-1, 1]. The cosine function is not one-to-one across its full domain. However, when the domain is restricted to [0, π], the function becomes one-to-one and so possesses an inverse function called the inverse cosine function or arc cosine function, written as cos⁻¹.

The definition is: y = cos⁻¹ x if and only if x = cos y and y is in [0, π].

Key properties follow:

(i) The domain of cos⁻¹ x is [-1, 1], and its range is [0, π].
(ii) cos(cos⁻¹ x) = x for x in [-1, 1], meaning |x| ≤ 1.
(iii) cos⁻¹(cos y) = y for y in [0, π].
(iv) The function cos⁻¹: [-1, 1] → [0, π] is strictly decreasing and is one-to-one.

The part of the curve where 0 ≤ y ≤ π is the principal value branch of y = cos⁻¹ x, and these y values are the principal values.

Note: The range of other branches of cos⁻¹ x are [π, 2π], [2π, 3π], [-π, 0], etc.

3. Inverse tangent function

Consider the tangent function f defined by f(x) = tan x. Its domain Df is all real numbers except odd multiples of π/2, and its range Rf = R. When we restrict the domain to the open interval (-π/2, π/2), the function becomes one-to-one, so it has an inverse function called the inverse tangent function or arc tangent function, denoted tan⁻¹.

This means: y = tan⁻¹ x if and only if x = tan y and y is in (-π/2, π/2).

Its key properties are:

(i) The domain of tan⁻¹ x is R, and its range is (-π/2, π/2).
(ii) tan(tan⁻¹ x) = x for all x in R.
(iii) tan⁻¹(tan y) = y for y in (-π/2, π/2).
(iv) The function tan⁻¹: R → (-π/2, π/2) is strictly increasing and is one-to-one.

The part where -π/2 < y < π/2 is called the principal value branch of y = tan⁻¹ x, and these y values are the principal values.

4. Inverse cotangent function

Consider the cotangent function f defined by f(x) = cot x. The domain Df is all real numbers except even multiples of π/2, and the range Rf = R. When we restrict the domain to the open interval (0, π), the function is one-to-one and has an inverse called the inverse cotangent function or arc cotangent function, written cot⁻¹.

This gives: y = cot⁻¹ x if and only if x = cot y and y is in (0, π).

Main properties include:

(i) The domain of cot⁻¹ x is R, and its range is (0, π).
(ii) cot(cot⁻¹ x) = x for all x in R.
(iii) cot⁻¹(cot y) = y for y in (0, π).
(iv) The function cot⁻¹: R → (0, π) is strictly decreasing and is one-to-one.

The region where 0 < y < π is the principal value branch of y = cot⁻¹ x, and these y values are called principal values.

5. Inverse secant function

Take the function f defined by f(x) = sec x. The domain Df includes all real numbers except odd multiples of π/2, and the range Rf = (-∞, -1] ∪ [1, ∞). By limiting the domain to [0, π/2) ∪ (π/2, π], the function becomes one-to-one and has an inverse function called the inverse secant function or arc secant function, denoted sec⁻¹.

We write: y = sec⁻¹ x if and only if x = sec y and y is in [0, π/2) ∪ (π/2, π].

It has these properties:

(i) The domain of sec⁻¹ x is (-∞, -1] ∪ [1, ∞), and its range is [0, π/2) ∪ (π/2, π].
(ii) sec(sec⁻¹ x) = x for |x| ≥ 1.
(iii) sec⁻¹(sec y) = y for y in (0, π) where y ≠ π/2.
(iv) sec⁻¹ x is strictly increasing in sections and is one-to-one.

The portion where 0 ≤ y ≤ π and y ≠ π/2 forms the principal value branch of y = sec⁻¹ x, and these y values are the principal values.

6. Inverse cosecant function

Consider the function f where f(x) = cosec x. The domain Df is all real numbers except even multiples of π/2, and the range Rf = (-∞, -1] ∪ [1, ∞). When the domain is limited to [-π/2, 0) ∪ (0, π/2], the function is one-to-one and has an inverse function called the inverse cosecant function or arc cosecant function, written cosec⁻¹.

The relationship is: y = cosec⁻¹ x if and only if x = cosec y, where y is in [-π/2, 0) ∪ (0, π/2].

Its characteristics are:

(i) The domain of cosec⁻¹ x is (-∞, -1] ∪ [1, ∞), and its range is [-π/2, 0) ∪ (0, π/2].
(ii) cosec(cosec⁻¹ x) = x for |x| ≥ 1.
(iii) cosec⁻¹(cosec y) = y for y in [-π/2, π/2] where y ≠ 0.
(iv) cosec⁻¹ x is strictly decreasing in sections and is one-to-one.

The part where -π/2 ≤ y ≤ π/2 and y ≠ 0 is known as the principal value branch of y = cosec⁻¹ x, and these y values are known as principal values.

 

5.3 Some Important Results

1. sin⁻¹ x + cos⁻¹ x = π/2, |x| ≤ 1.

2. tan⁻¹ x + cot⁻¹ x = π/2, x ∈ R.

3. sec⁻¹ x + cosec⁻¹ x = π/2, |x| ≥ 1.

4. tan⁻¹ x + tan⁻¹ y = tan⁻¹((x + y)/(1 - xy)), xy < 1.

5. tan⁻¹ x - tan⁻¹ y = tan⁻¹((x - y)/(1 + xy)), xy > -1.

6. (i) sin⁻¹ x = cos⁻¹(√(1 - x²)), 0 ≤ x ≤ 1
(ii) cos⁻¹ x = sin⁻¹(√(1 - x²)), 0 ≤ x ≤ 1.

7. (i) tan⁻¹(x/√(1 - x²)) = sin⁻¹ x, |x| < 1
(ii) tan⁻¹(√(1 - x²)/x) = cos⁻¹ x, 0 < x ≤ 1.

8. sin⁻¹(2x/(1 + x²)) = 2 tan⁻¹ x, |x| ≤ 1.

9. cos⁻¹((1 - x²)/(1 + x²)) = 2 tan⁻¹ x, x ≥ 0.

10. tan⁻¹(2x/(1 - x²)) = 2 tan⁻¹ x, |x| < 1.

11. cosec⁻¹ x = sin⁻¹(1/x), |x| ≥ 1.

12. sec⁻¹ x = cos⁻¹(1/x), |x| ≥ 1.

13. cot⁻¹ x = tan⁻¹(1/x) when x > 0; cot⁻¹ x = π + tan⁻¹(1/x) when x < 0.

14. cos(sin⁻¹ x) = sin(cos⁻¹ x) = √(1 - x²), |x| ≤ 1.

15. (i) sin⁻¹(-x) = -sin⁻¹ x, |x| ≤ 1
(ii) cos⁻¹(-x) = π - cos⁻¹ x, |x| ≤ 1
(iii) tan⁻¹(-x) = -tan⁻¹ x, x ∈ R
(iv) cot⁻¹(-x) = π - cot⁻¹ x, x ∈ R
(v) cosec⁻¹(-x) = -cosec⁻¹ x, |x| ≥ 1
(vi) sec⁻¹(-x) = π - sec⁻¹ x, |x| ≥ 1.

16. For suitable values of x and y:

(i) sin⁻¹ x + sin⁻¹ y = sin⁻¹(x√(1 - y²) + y√(1 - x²))
(ii) sin⁻¹ x - sin⁻¹ y = sin⁻¹(x√(1 - y²) - y√(1 - x²))
(iii) cos⁻¹ x + cos⁻¹ y = cos⁻¹(xy - √(1 - x²)√(1 - y²))
(iv) cos⁻¹ x - cos⁻¹ y = cos⁻¹(xy + √(1 - x²)√(1 - y²)).

Proof 1. Let sin⁻¹ x = y, so x = sin y. We can write x = cos(π/2 - y). Therefore cos⁻¹ x = π/2 - y, which means cos⁻¹ x = π/2 - sin⁻¹ x. Adding sin⁻¹ x to both sides gives sin⁻¹ x + cos⁻¹ x = π/2 for |x| ≤ 1. (When y = sin⁻¹ x is a principal value, we have -π/2 ≤ y ≤ π/2. This means π/2 ≥ -y ≥ -π/2. Adding π/2 to the middle term gives π/2 + π/2 ≥ π/2 - y ≥ π/2 - π/2, so π ≥ π/2 - y ≥ 0, meaning 0 ≤ π/2 - y ≤ π. Thus π/2 - y is a principal value of cos⁻¹ x.)

Proof 2. Let tan⁻¹ x = y, so x = tan y. We can write x = cot(π/2 - y). Therefore cot⁻¹ x = π/2 - y, which gives cot⁻¹ x = π/2 - tan⁻¹ x. This leads to tan⁻¹ x + cot⁻¹ x = π/2.

Proof 3. Left as an exercise for the reader.

Proof 4. Let tan⁻¹ x = α and tan⁻¹ y = β, so x = tan α and y = tan β. Now tan(α + β) = (tan α + tan β)/(1 - tan α tan β) = (x + y)/(1 - xy). Therefore α + β = tan⁻¹((x + y)/(1 - xy)), which means tan⁻¹ x + tan⁻¹ y = tan⁻¹((x + y)/(1 - xy)). Note that when α = tan⁻¹ x and β = tan⁻¹ y, we have -π/2 < α < π/2 and -π/2 < β < π/2. Given xy < 1, we get tan α tan β < 1, so (sin α sin β)/(cos α cos β) < 1. Since cos α cos β > 0 (because cos α > 0 and cos β > 0 when -π/2 < α < π/2 and -π/2 < β < π/2), we have sin α sin β < cos α cos β. Thus 0 < cos α cos β - sin α sin β, meaning 0 < cos(α + β). This gives -π/2 < α + β < π/2.

Proof 5. Left as an exercise for the reader.

Proof 6 (i). Let sin⁻¹ x = y, so x = sin y. Since 0 ≤ x ≤ 1, we have 0 ≤ y ≤ π/2. Therefore √(1 - x²) = √(1 - sin² y) = √(cos² y) = |cos y| = cos y (since 0 ≤ y ≤ π/2 means cos y ≥ 0, so |cos y| = cos y). Thus y = cos⁻¹(√(1 - x²)), which gives sin⁻¹ x = cos⁻¹(√(1 - x²)).

Proof 6 (ii). Let cos⁻¹ x = y, so x = cos y. Since 0 ≤ x ≤ 1, we have 0 ≤ y ≤ π/2. Therefore √(1 - x²) = √(1 - cos² y) = √(sin² y) = |sin y| = sin y (since 0 ≤ y ≤ π/2 means sin y ≥ 0, so |sin y| = sin y). Thus y = sin⁻¹(√(1 - x²)), which gives cos⁻¹ x = sin⁻¹(√(1 - x²)).

Proof 7 (i). Let x = sin y, so y = sin⁻¹ x with -π/2 < y < π/2 (since |x| < 1). Then √(1 - x²) = √(1 - sin² y) = √(cos² y) = |cos y| = cos y (because -π/2 < y < π/2 means cos y > 0). Therefore tan⁻¹(x/√(1 - x²)) = tan⁻¹(sin y/cos y) = tan⁻¹(tan y) = y = sin⁻¹ x.

Proof 7 (ii). Let x = cos y, so y = cos⁻¹ x with 0 ≤ y < π/2 (since 0 < x ≤ 1). Then √(1 - x²) = √(1 - cos² y) = √(sin² y) = |sin y| = sin y (since 0 ≤ y < π/2 means sin y ≥ 0). Therefore tan⁻¹(√(1 - x²)/x) = tan⁻¹(sin y/cos y) = tan⁻¹(tan y) = y = cos⁻¹ x.

Proof 8. Let tan⁻¹ x = y, so x = tan y with |x| ≤ 1 and -π/4 ≤ y ≤ π/4. Now sin 2y = 2 tan y/(1 - tan² y) = 2x/(1 - x²). Therefore 2y = sin⁻¹(2x/(1 - x²)) (since -π/4 ≤ y ≤ π/4 means -π/2 ≤ 2y ≤ π/2). Thus 2 tan⁻¹ x = sin⁻¹(2x/(1 - x²)).

Proof 9. Left as an exercise for the reader.

Proof 10. Using result 4 with y = x, we get tan⁻¹ x + tan⁻¹ x = tan⁻¹((x + x)/(1 - x·x)), so 2 tan⁻¹ x = tan⁻¹(2x/(1 - x²)).

Proof 11. Let cosec⁻¹ x = y with -π/2 ≤ y ≤ π/2 and y ≠ 0. Then x = cosec y, so 1/x = 1/cosec y = sin y. Therefore y = sin⁻¹(1/x) (since -π/2 ≤ y ≤ π/2 and y ≠ 0). Thus cosec⁻¹ x = sin⁻¹(1/x) for |x| ≥ 1.

Proof 12. Let sec⁻¹ x = y with 0 ≤ y ≤ π and y ≠ π/2. Then x = sec y, so 1/x = 1/sec y = cos y. Therefore y = cos⁻¹(1/x) (since 0 ≤ y ≤ π and y ≠ π/2). Thus sec⁻¹ x = cos⁻¹(1/x) for |x| ≥ 1.

Proof 13 (When x > 0): Let cot⁻¹ x = y with 0 < y < π/2 (since x > 0). Then x = cot y, so 1/x = 1/cot y = tan y. Therefore y = tan⁻¹(1/x) (since 0 < y < π/2). Thus cot⁻¹ x = tan⁻¹(1/x) for x > 0.

When x < 0: Let cot⁻¹ x = y with π/2 < y < π. Now π/2 < y < π means -π/2 < y - π < 0. From cot⁻¹ x = y, we get x = cot y, so 1/x = cot y = -tan(π - y) = tan(y - π). Therefore y - π = tan⁻¹(1/x) (since -π/2 < y - π < 0). Thus cot⁻¹ x = π + tan⁻¹(1/x) for x < 0.

Proof 14. Let sin⁻¹ x = y with -π/2 ≤ y ≤ π/2. Then x = sin y, and for this value of y, cos y ≥ 0. So cos y = √(1 - sin² y) = √(1 - x²), giving cos(sin⁻¹ x) = √(1 - x²).

Now let cos⁻¹ x = t with 0 ≤ t ≤ π. Then x = cos t, and for this value of t, sin t ≥ 0. So sin t = √(1 - cos² t) = √(1 - x²), giving sin(cos⁻¹ x) = √(1 - x²).

From both results, cos(sin⁻¹ x) = sin(cos⁻¹ x) = √(1 - x²) for |x| ≤ 1.

Proof 15 (i). Let sin⁻¹ x = y with -π/2 ≤ y ≤ π/2. Then x = sin y, so -x = -sin y. Rearranging gives -x = sin(-y) with -π/2 ≤ -y ≤ π/2. Thus sin⁻¹(-x) = -y = -sin⁻¹ x for |x| ≤ 1.

Proof 15 (ii). Let cos⁻¹ x = y with 0 ≤ y ≤ π. Then x = cos y, so -x = -cos y. Since 0 ≥ -y ≥ -π, we have π ≥ π - y ≥ 0. From -x = cos(π - y) with 0 ≤ π - y ≤ π, we get cos⁻¹(-x) = π - y = π - cos⁻¹ x for |x| ≤ 1.

Proof 15 (iii). The proofs of the other parts are left as exercises for the reader.

Proof 16 (i). Let sin⁻¹ x = α and sin⁻¹ y = β, so x = sin α and y = sin β. Then cos α = √(1 - sin² α) = √(1 - x²) and cos β = √(1 - sin² β) = √(1 - y²). Now sin(α + β) = sin α cos β + cos α sin β = x√(1 - y²) + √(1 - x²)y. Therefore α + β = sin⁻¹(x√(1 - y²) + y√(1 - x²)), which means sin⁻¹ x + sin⁻¹ y = sin⁻¹(x√(1 - y²) + y√(1 - x²)). Note: Here |x| ≤ 1 and |y| ≤ 1 if xy ≤ 0 or if xy > 0 and x² + y² ≤ 1.

Proof 16 (ii). Left as an exercise for the reader.

Proof 16 (iii). Let cos⁻¹ x = α and cos⁻¹ y = β, so x = cos α and y = cos β. Then sin α = √(1 - cos² α) = √(1 - x²) and sin β = √(1 - cos² β) = √(1 - y²). Now cos(α + β) = cos α cos β - sin α sin β = xy - √(1 - x²)√(1 - y²). Therefore α + β = cos⁻¹(xy - √(1 - x²)√(1 - y²)), which means cos⁻¹ x + cos⁻¹ y = cos⁻¹(xy - √(1 - x²)√(1 - y²)).

Proof 16 (iv). Left as an exercise for the reader.

 

Illustrative Examples

 

Example 1. Find the principal values of: (i) cos⁻¹(√3/2) (ii) cot⁻¹(√3) (iii) cosec⁻¹(√2)
Answer:
(i) Let cos⁻¹(√3/2) = x where 0 ≤ x ≤ π. Then cos x = √3/2 = cos(π/6), so x = π/6. Therefore cos⁻¹(√3/2) = π/6.
(ii) Let cot⁻¹(√3) = x where 0 < x < π. Then cot x = √3 = cot(π/6), so x = π/6. Therefore cot⁻¹(√3) = π/6.
(iii) Let cosec⁻¹(√2) = x where -π/2 ≤ x ≤ π/2 and x ≠ 0. Then cosec x = √2 = cosec(π/4), so x = π/4. Therefore cosec⁻¹(√2) = π/4.
In simple words: Use the definition of each inverse function to set up an equation. Find the angle whose trigonometric value matches the given number, making sure it falls in the principal range.

Exam Tip: Always check that your answer lies within the specified range for each inverse function - this is crucial for getting the principal value.

 

Example 2. Find the principal values of: (i) sin⁻¹(-√3/2) (ii) sec⁻¹(-2) (iii) cot⁻¹(-1)
Answer:
(i) Let sin⁻¹(-√3/2) = x where -π/2 ≤ x ≤ π/2. Then sin x = -√3/2 = -sin(π/3) = sin(-π/3), so x = -π/3. Therefore sin⁻¹(-√3/2) = -π/3.

Alternative method: Using sin⁻¹(-x) = -sin⁻¹(x), we get sin⁻¹(-√3/2) = -sin⁻¹(√3/2). Let sin⁻¹(√3/2) = x where -π/2 ≤ x ≤ π/2. Then sin x = √3/2 = sin(π/3), so x = π/3. Thus sin⁻¹(√3/2) = π/3, and sin⁻¹(-√3/2) = -π/3.

(ii) Using sec⁻¹(-x) = π - sec⁻¹(x), we get sec⁻¹(-2) = π - sec⁻¹(2). Let sec⁻¹(2) = x where 0 ≤ x ≤ π and x ≠ π/2. Then sec x = 2 = sec(π/3), so x = π/3. Thus sec⁻¹(2) = π/3, and sec⁻¹(-2) = π - π/3 = 2π/3.

(iii) Using cot⁻¹(-x) = π - cot⁻¹(x), we get cot⁻¹(-1) = π - cot⁻¹(1). Let cot⁻¹(1) = x where 0 < x < π. Then cot x = 1 = cot(π/4), so x = π/4. Thus cot⁻¹(1) = π/4, and cot⁻¹(-1) = π - π/4 = 3π/4.
In simple words: When the input is negative, use the given property (like sin⁻¹(-x) = -sin⁻¹(x) or cos⁻¹(-x) = π - cos⁻¹(x)) to work with the positive version first, then apply the transformation.

Exam Tip: Memorize the negative-input properties for each inverse function - they save time and reduce calculation errors.

 

Example 3. Find the principal values of: (i) sin⁻¹(sin(3π/5)) (ii) cos⁻¹(cos(7π/6)) (iii) tan⁻¹(tan(6π/7))
Answer:
(i) Let sin⁻¹(sin(3π/5)) = y where -π/2 ≤ y ≤ π/2. Then sin y = sin(3π/5). Since 3π/5 is outside the range [-π/2, π/2], we rewrite: sin(3π/5) = sin(π - 2π/5) = sin(2π/5), which is in the acceptable range. So y = 2π/5, giving sin⁻¹(sin(3π/5)) = 2π/5.

(ii) Let cos⁻¹(cos(7π/6)) = y where 0 ≤ y ≤ π. Then cos y = cos(7π/6). Since 7π/6 is outside the range [0, π], we rewrite: cos(7π/6) = cos(2π - 5π/6) = cos(5π/6), which is in the acceptable range. So y = 5π/6, giving cos⁻¹(cos(7π/6)) = 5π/6.

(iii) Let tan⁻¹(tan(6π/7)) = y where -π/2 < y < π/2. Then tan y = tan(6π/7). Since 6π/7 is outside the range (-π/2, π/2), we rewrite: tan(6π/7) = tan(π - π/7) = -tan(π/7) = tan(-π/7), which is in the acceptable range. So y = -π/7, giving tan⁻¹(tan(6π/7)) = -π/7.
In simple words: The angle you are given may be outside the principal range. Use trigonometric identities to find an equivalent angle that lies in the correct range, then apply the inverse function.

Exam Tip: Learn reduction formulas like sin(π - θ) = sin(θ), cos(2π - θ) = cos(θ), and tan(π + θ) = tan(θ) - they help shift angles into the principal range quickly.

 

Example 4. Show that sin⁻¹(√3/2) + 2 tan⁻¹(1/√3) = 2π/3
Answer: Let sin⁻¹(√3/2) = α where -π/2 ≤ α ≤ π/2. Then sin α = √3/2 = sin(π/3), so α = π/3.

Let tan⁻¹(1/√3) = β where -π/2 < β < π/2. Then tan β = 1/√3 = tan(π/6), so β = π/6. Therefore 2 tan⁻¹(1/√3) = 2β = π/3.

Adding both: sin⁻¹(√3/2) + 2 tan⁻¹(1/√3) = π/3 + π/3 = 2π/3.
In simple words: Find each inverse separately and simplify to its principal value. Then do the arithmetic to verify the equation.

Exam Tip: Break multi-step problems into smaller pieces - find each inverse function value, then combine them using basic arithmetic or identities.

 

Example 33. If sin⁻¹ x + sin⁻¹ y + sin⁻¹ z = π, prove that: (i) x² - y² - z² + 2yz√(1 - x²) = 0 (ii) x⁴ + y⁴ + z⁴ + 4x²y²z² = 2(x²y² + y²z² + z²x²)
Answer:
(i) Starting with sin⁻¹ x + sin⁻¹ y + sin⁻¹ z = π, we rearrange to get sin⁻¹ x + sin⁻¹ y = π - sin⁻¹ z. Using the addition formula for inverse sine on the left side: sin⁻¹(x√(1 - y²) + y√(1 - x²)) = π - sin⁻¹ z. Since sin(π - θ) = sin(θ), we have sin⁻¹(x√(1 - y²) + y√(1 - x²)) = sin⁻¹(z). Therefore x√(1 - y²) + y√(1 - x²) = z, which gives x√(1 - y²) = z - y√(1 - x²). Squaring both sides: x²(1 - y²) = z² + y²(1 - x²) - 2yz√(1 - x²). Expanding: x² - x²y² = z² + y² - x²y² - 2yz√(1 - x²). Simplifying yields x² - y² - z² + 2yz√(1 - x²) = 0.

(ii) From part (i), we have x² - y² - z² = -2yz√(1 - x²). Squaring: (x² - y² - z²)² = 4y²z²(1 - x²). Expanding the left side: x⁴ + y⁴ + z⁴ - 2x²y² - 2z²x² + 2y²z² = 4y²z² - 4x²y²z². Rearranging: x⁴ + y⁴ + z⁴ + 4x²y²z² = 2(x²y² + y²z² + z²x²).
In simple words: Start by using the condition given. Apply inverse function addition properties to get a simpler form. Square equations strategically to eliminate square roots, then expand and collect like terms.

Exam Tip: When a condition involves multiple inverse functions, use the appropriate addition/subtraction formula to combine them. Square carefully to avoid introducing extraneous solutions.

 

Exercise 5.1

 

Question 1. Find the principal values of:
(i) sin⁻¹(1/√2)
(ii) cos⁻¹(-1/2)
(iii) cot⁻¹(-√3)
(iv) tan⁻¹(-1/√3)
(v) cosec⁻¹(-2)
(vi) sec⁻¹(2/√3)
Answer: (i) π/4, (ii) 2π/3, (iii) 2π/3, (iv) -π/6, (v) -π/6, (vi) π/6
In simple words: Find the angle in the principal range whose trigonometric value matches the input number.

Exam Tip: Check the range for each inverse function carefully - sin⁻¹ and tan⁻¹ include negative angles, while cos⁻¹ and cot⁻¹ do not.

 

Question 2. Evaluate the following:
(i) sin⁻¹(sin(5π/6))
(ii) tan⁻¹(sin(-π/2))
(iii) tan⁻¹(tan(3π/4))
(iv) cot(tan⁻¹(√3))
(v) sin(π/6 - sin⁻¹(-√3/2))
(vi) cos(cos⁻¹(-√3/2) + π/6)
Answer: (i) π/6, (ii) -1, (iii) -π/4, (iv) 1/√3, (v) 1/2, (vi) 1/2
In simple words: Apply inverse function properties and simplify step-by-step, converting outside angles to the principal range where needed.

Exam Tip: Work from the inside out - evaluate the innermost expression first, then apply the outer function.

 

Question 3. Show that:
(i) tan⁻¹(tan(5π/6)) ≠ 5π/6. What is its value?
(ii) cos⁻¹(cos(-π/6)) ≠ -π/6. What is its value?
(iii) sin⁻¹(sin(2π/3)) ≠ 2π/3. What is its value?
Answer: (i) The value is -π/6 (the given angle 5π/6 is outside the range (-π/2, π/2)). (ii) The value is π/6 (the given angle -π/6 is outside the range [0, π]). (iii) The value is π/3 (the given angle 2π/3 is outside the range [-π/2, π/2]).
In simple words: When the argument is outside the principal range, use a trigonometric identity to find an equivalent angle inside that range, then apply the inverse function.

Exam Tip: These problems highlight a key point: \( f^{-1}(f(x)) = x \) only when x is in the principal range of \( f^{-1} \).

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