Inverse Trigonometric Functions and Their Properties

In trigonometry, we deal with the relationships between angles and sides of triangles. Inverse trigonometric functions are the inverse operations of the standard trigonometric functions. Just as subtraction is the inverse of addition, and division is the inverse of multiplication, inverse trigonometric functions "undo" the work of the original trigonometric functions.

For example, if sin(θ) = x, then θ = sin-1(x). The function sin-1 is called the inverse sine function, or arcsine function, denoted as arcsin(x). Similarly, we have inverse cosine (arccos or cos-1), inverse tangent (arctan or tan-1), and so on for cotangent, secant, and cosecant.

Definition of Inverse Trigonometric Functions

A function has an inverse if and only if it is one-to-one (injective) and onto (surjective). Standard trigonometric functions like sin(x), cos(x), and tan(x) are periodic, meaning they repeat their values over intervals. This periodicity makes them not one-to-one over their entire domain. To define their inverse functions, we restrict their domains to specific intervals where they are one-to-one.

1. Inverse Sine Function (arcsin x or sin^-1 x)

The function y = sin(x) is one-to-one in the interval [-π/2, π/2]. The range of sin(x) in this interval is [-1, 1]. Therefore, the inverse sine function, y = sin-1(x), is defined as:

• Domain: [-1, 1]
• Range: [-π/2, π/2]

This means that for any value x between -1 and 1, sin-1(x) will give an angle y between -π/2 and π/2 such that sin(y) = x.

2. Inverse Cosine Function (arccos x or cos^-1 x)

The function y = cos(x) is one-to-one in the interval [0, π]. The range of cos(x) in this interval is [-1, 1]. Therefore, the inverse cosine function, y = cos-1(x), is defined as:

• Domain: [-1, 1]
• Range: [0, π]

For any value x between -1 and 1, cos-1(x) will give an angle y between 0 and π such that cos(y) = x.

3. Inverse Tangent Function (arctan x or tan^-1 x)

The function y = tan(x) is one-to-one in the interval (-π/2, π/2). The range of tan(x) in this interval is (-∞, ∞). Therefore, the inverse tangent function, y = tan-1(x), is defined as:

• Domain: (-∞, ∞)
• Range: (-π/2, π/2)

For any real number x, tan-1(x) will give an angle y between -π/2 and π/2 such that tan(y) = x.

4. Inverse Cotangent Function (arccot x or cot^-1 x)

The function y = cot(x) is one-to-one in the interval (0, π). The range of cot(x) in this interval is (-∞, ∞). Therefore, the inverse cotangent function, y = cot-1(x), is defined as:

• Domain: (-∞, ∞)
• Range: (0, π)

For any real number x, cot-1(x) will give an angle y between 0 and π such that cot(y) = x.

5. Inverse Secant Function (arcsec x or sec^-1 x)

The function y = sec(x) is one-to-one in the intervals [0, π/2) and (π/2, π]. Conventionally, its inverse, y = sec-1(x), is defined with the range [0, π/2) ∪ (π/2, π]. The domain is (-∞, -1] ∪ [1, ∞).

• Domain: (-∞, -1] ∪ [1, ∞)
• Range: [0, π/2) ∪ (π/2, π]

For any x such that |x| ≥ 1, sec-1(x) gives an angle y in [0, π/2) ∪ (π/2, π] such that sec(y) = x.

6. Inverse Cosecant Function (arccsc x or csc^-1 x)

The function y = csc(x) is one-to-one in the intervals [-π/2, 0) and (0, π/2]. Conventionally, its inverse, y = csc-1(x), is defined with the range [-π/2, 0) ∪ (0, π/2]. The domain is (-∞, -1] ∪ [1, ∞).

• Domain: (-∞, -1] ∪ [1, ∞)
• Range: [-π/2, 0) ∪ (0, π/2]

For any x such that |x| ≥ 1, csc-1(x) gives an angle y in [-π/2, 0) ∪ (0, π/2] such that csc(y) = x.

Properties of Inverse Trigonometric Functions

Inverse trigonometric functions have several important properties that are useful in simplifying expressions and solving problems. These properties stem from the relationships between the original trigonometric functions and their inverses, as well as the definitions of the inverse functions themselves.

1. Reciprocal Properties

These properties relate an inverse trigonometric function of x to the inverse trigonometric function of 1/x.

• sin-1(x) = csc-1(1/x), for |x| ≥ 1
• cos-1(x) = sec-1(1/x), for |x| ≥ 1
• tan-1(x) = cot-1(1/x), for x > 0
• tan-1(x) = π + cot-1(1/x), for x < 0
• cot-1(x) = tan-1(1/x), for x > 0
• cot-1(x) = π + tan-1(1/x), for x < 0
• sec-1(x) = cos-1(1/x), for |x| ≥ 1
• csc-1(x) = sin-1(1/x), for |x| ≥ 1

Let's prove the property tan-1(x) = cot-1(1/x) for x > 0. Let y = tan-1(x). Since x > 0, y is in the interval (0, π/2). This means tan(y) = x. Since tan(y) = x, we have cot(π/2 - y) = x. Also, cot(y) = 1/x. If we consider cot-1(1/x), we need its value to be in the range (0, π). Since tan(y) = x, then cot(y) = 1/x. If x > 0, then y ∈ (0, π/2). Therefore, 1/x > 0. And cot(y) = 1/x. The principal range for cot-1 is (0, π). Since y ∈ (0, π/2), y is already in the range of cot-1. So, cot-1(1/x) = y. Substituting back y = tan-1(x), we get tan-1(x) = cot-1(1/x) for x > 0.

For x < 0, let y = tan-1(x). Then y ∈ (-π/2, 0). tan(y) = x. cot(y) = 1/x. Since y ∈ (-π/2, 0), cot(y) is negative. We know cot(π + z) = cot(z). Let z = y. So cot(π + y) = cot(y) = 1/x. Since y ∈ (-π/2, 0), then π + y ∈ (π/2, π). The value π + y lies in the range of cot-1, which is (0, π). So, cot-1(1/x) = π + y. Substituting y = tan-1(x), we get cot-1(1/x) = π + tan-1(x). Rearranging, tan-1(x) = cot-1(1/x) - π. Wait, there was a sign error in the property statement above. Let's re-derive. Let y = tan-1(x). If x < 0, then y ∈ (-π/2, 0). So, tan(y) = x. Then cot(y) = 1/x. We want to express this in terms of cot-1(1/x). The range of cot-1 is (0, π). Since y ∈ (-π/2, 0), we can write y = - (π/2 - y) if we want a positive angle. This is not helpful. Let's use the identity cot(θ) = -cot(-θ). We have cot(y) = 1/x. Consider cot(π + y) = cot(y) = 1/x. Since y ∈ (-π/2, 0), then π + y ∈ (π/2, π). This is in the range of cot-1. So, cot-1(1/x) = π + y. Substituting y = tan-1(x), we get cot-1(1/x) = π + tan-1(x). This implies tan-1(x) = cot-1(1/x) - π. Let's check the original property statement again. Ah, the property tan-1(x) = π + cot-1(1/x) for x < 0 is correct. Let's see why. Let y = tan-1(x). For x < 0, y ∈ (-π/2, 0). So tan(y) = x. Then cot(y) = 1/x. We want to relate this to cot-1(1/x). Consider cot(π - y) = -cot(y) = -1/x. This is not 1/x. Let's use the relation cot(θ) = tan(π/2 - θ). Let y = tan-1(x). So tan(y) = x. Consider cot-1(1/x). Let this be α. So cot(α) = 1/x. Since x < 0, 1/x < 0. The range of cot-1 is (0, π). So α ∈ (π/2, π). We have cot(α) = 1/x. This means tan(α) = x. So we have tan(y) = x and tan(α) = x. Since y ∈ (-π/2, 0) and α ∈ (π/2, π), they are not equal. We know that tan(θ) = tan(θ + π). So, tan(α) = tan(α - π). Since α ∈ (π/2, π), then α - π ∈ (-π/2, 0). This interval (-π/2, 0) is the range of tan-1 for negative values. So, tan-1(x) = α - π. Substituting α = cot-1(1/x), we get tan-1(x) = cot-1(1/x) - π. This matches my derivation. Let me double check the standard properties. Ah, the property is indeed tan-1(x) = π + cot-1(1/x) for x < 0. Let's try to prove this one. Let y = tan-1(x). For x < 0, y ∈ (-π/2, 0). So, tan(y) = x. We know that cot(π/2 + y) = -tan(y) = -x. This is not useful. Let's use cot(θ) = tan(π/2 - θ). We have tan(y) = x. Consider cot-1(1/x). Let this be α. So cot(α) = 1/x. Since x < 0, 1/x < 0. So α ∈ (π/2, π). From cot(α) = 1/x, we get tan(α) = x. We have tan(y) = x and tan(α) = x. Since y ∈ (-π/2, 0) and α ∈ (π/2, π), we need to relate them. We know tan(θ) = tan(θ + nπ) for integer n. So, tan(α) = tan(α - π). Since α ∈ (π/2, π), α - π ∈ (-π/2, 0). This interval is the range for tan-1(x) when x < 0. So, tan-1(x) = α - π. Substituting α = cot-1(1/x), we get tan-1(x) = cot-1(1/x) - π. This is consistent. The property statement in many textbooks is tan-1(x) = cot-1(1/x) - π for x < 0. Let me re-check the provided property: tan-1(x) = π + cot-1(1/x) for x < 0. This means cot-1(1/x) = tan-1(x) - π. This contradicts my derivation. Let's try a numerical example. Let x = -1. tan-1(-1) = -π/4. 1/x = -1. cot-1(-1) = 3π/4 (since cot(3π/4) = -1 and 3π/4 ∈ (0, π)). According to the property tan-1(x) = π + cot-1(1/x): -π/4 = π + 3π/4 = 7π/4. This is false. According to my derived property tan-1(x) = cot-1(1/x) - π: -π/4 = 3π/4 - π = -π/4. This is true. Okay, so the property should be tan-1(x) = cot-1(1/x) - π for x < 0. And cot-1(x) = tan-1(1/x) + π for x < 0. Let's check this one. Let x = -1. cot-1(-1) = 3π/4. 1/x = -1. tan-1(-1) = -π/4. Property: 3π/4 = -π/4 + π = 3π/4. This is true. So the properties for negative values are: * tan-1(x) = cot-1(1/x) - π, for x < 0 * cot-1(x) = tan-1(1/x) + π, for x < 0 * sec-1(x) = π - cos-1(1/x), for x ≤ -1 (Proof: Let y = sec-1(x). Then y ∈ (π/2, π]. sec(y) = x. cos(y) = 1/x. Let α = cos-1(1/x). Then α ∈ [π/2, π). cos(α) = 1/x. So cos(y) = cos(α). Since y and α are in the same interval (π/2, π], we have y = α. So sec-1(x) = cos-1(1/x). This seems to contradict the property. Let's recheck the range of sec-1(x) for x ≤ -1. It is (π/2, π]. The range of cos-1(1/x) for x ≤ -1 is when 1/x ∈ [-1, 0). The range of cos-1 for [-1, 0) is (π/2, π]. So they match. Thus, sec-1(x) = cos-1(1/x) for x ≤ -1. The property sec-1(x) = π - cos-1(1/x) is for the range of sec-1(x) being [0, π/2) ∪ (π/2, π]. Let's assume the standard definition of sec-1(x) where the range is [0, π/2) ∪ (π/2, π]. If x ≤ -1, then 1/x ∈ [-1, 0). Let y = sec-1(x). Then y ∈ (π/2, π]. sec(y) = x. cos(y) = 1/x. Let α = cos-1(1/x). Since 1/x ∈ [-1, 0), α ∈ (π/2, π]. So cos(α) = 1/x. Thus, cos(y) = cos(α). Since both y and α are in the interval (π/2, π], we have y = α. Therefore, sec-1(x) = cos-1(1/x) for x ≤ -1. The property sec-1(x) = π - cos-1(1/x) is likely incorrect or for a different definition. Let's stick to the simpler derived one. * csc-1(x) = π - sin-1(1/x), for x ≤ -1 (Similar logic, the range of csc-1(x) for x ≤ -1 is [-π/2, 0). The range of sin-1(1/x) for x ≤ -1 is [-π/2, 0). Let y = csc-1(x). y ∈ [-π/2, 0). csc(y) = x. sin(y) = 1/x. Let α = sin-1(1/x). Since 1/x ∈ [-1, 0), α ∈ [-π/2, 0). So sin(α) = 1/x. Thus sin(y) = sin(α). Since both y and α are in [-π/2, 0), we have y = α. Therefore, csc-1(x) = sin-1(1/x) for x ≤ -1.

Let's summarize the corrected reciprocal properties:

• sin-1(x) = csc-1(1/x), for |x| ≥ 1
• cos-1(x) = sec-1(1/x), for |x| ≥ 1
• tan-1(x) = cot-1(1/x), for x > 0
• tan-1(x) = cot-1(1/x) - π, for x < 0
• cot-1(x) = tan-1(1/x) + π, for x < 0
• sec-1(x) = cos-1(1/x), for |x| ≥ 1
• csc-1(x) = sin-1(1/x), for |x| ≥ 1

2. Negative Argument Properties

These properties relate the inverse trigonometric function of -x to the inverse trigonometric function of x.

• sin-1(-x) = -sin-1(x), for x ∈ [-1, 1] (Odd function)
• tan-1(-x) = -tan-1(x), for all real x (Odd function)
• csc-1(-x) = -csc-1(x), for |x| ≥ 1 (Odd function)
• cos-1(-x) = π - cos-1(x), for x ∈ [-1, 1]
• cot-1(-x) = π - cot-1(x), for all real x
• sec-1(-x) = π - sec-1(x), for |x| ≥ 1

Let's prove cos-1(-x) = π - cos-1(x). Let y = cos-1(-x). Since x ∈ [-1, 1], -x ∈ [-1, 1]. The range of cos-1 is [0, π]. So y ∈ [0, π]. cos(y) = -x. We know that cos(π - θ) = -cos(θ). So, cos(π - y) = -cos(y) = -(-x) = x. Let α = cos-1(x). Since x ∈ [-1, 1], α ∈ [0, π]. We have cos(π - y) = x and cos(α) = x. If α ∈ [0, π], then π - α ∈ [0, π]. So, we have cos(π - y) = cos(α), where both π - y and α are in the range [0, π]. This implies π - y = α. Rearranging, y = π - α. Substituting back, cos-1(-x) = π - cos-1(x).

Similarly, for cot-1(-x) = π - cot-1(x). Let y = cot-1(-x). Then y ∈ (0, π). cot(y) = -x. We know cot(π - θ) = -cot(θ). So, cot(π - y) = -cot(y) = -(-x) = x. Let α = cot-1(x). Then α ∈ (0, π). cot(α) = x. We have cot(π - y) = x and cot(α) = x. If α ∈ (0, π), then π - α ∈ (0, π). So, cot(π - y) = cot(α), where both π - y and α are in the range (0, π). This implies π - y = α. Rearranging, y = π - α. Substituting back, cot-1(-x) = π - cot-1(x).

For sec-1(-x) = π - sec-1(x). Let y = sec-1(-x). Then y ∈ [0, π/2) ∪ (π/2, π]. sec(y) = -x. So cos(y) = -1/x. Let α = sec-1(x). Then α ∈ [0, π/2) ∪ (π/2, π]. sec(α) = x. So cos(α) = 1/x. We have cos(y) = -1/x = -cos(α). We know cos(π - θ) = -cos(θ). So, cos(π - α) = -cos(α) = -1/x. Thus, cos(y) = cos(π - α). We need to ensure that π - α is in the range of sec-1(-x). If α ∈ [0, π/2), then π - α ∈ (π/2, π]. This is in the range of sec-1(-x). If α ∈ (π/2, π], then π - α ∈ [0, π/2). This is also in the range of sec-1(-x). So, we can equate them: y = π - α. Substituting back, sec-1(-x) = π - sec-1(x).

3. Properties involving Sums and Differences of Angles

These are crucial for simplifying complex expressions.

• sin-1(x) + cos-1(x) = π/2, for x ∈ [-1, 1]
• tan-1(x) + cot-1(x) = π/2, for all real x
• sec-1(x) + csc-1(x) = π/2, for |x| ≥ 1

Let's prove sin-1(x) + cos-1(x) = π/2. Let y = sin-1(x). Then y ∈ [-π/2, π/2] and sin(y) = x. We know that sin(y) = cos(π/2 - y). So, x = cos(π/2 - y). Now, we need to find the range of π/2 - y. Since y ∈ [-π/2, π/2], then -y ∈ [-π/2, π/2]. And π/2 - y ∈ [0, π]. This interval [0, π] is the principal range of the cos-1 function. Since x = cos(π/2 - y) and π/2 - y ∈ [0, π], we can write: cos-1(x) = π/2 - y. Substituting y = sin-1(x), we get: cos-1(x) = π/2 - sin-1(x). Rearranging gives sin-1(x) + cos-1(x) = π/2.

Let's prove tan-1(x) + cot-1(x) = π/2. Let y = tan-1(x). Then y ∈ (-π/2, π/2) and tan(y) = x. We know that tan(y) = cot(π/2 - y). So, x = cot(π/2 - y). Now, find the range of π/2 - y. Since y ∈ (-π/2, π/2), then -y ∈ (-π/2, π/2). And π/2 - y ∈ (0, π). This interval (0, π) is the principal range of the cot-1 function. Since x = cot(π/2 - y) and π/2 - y ∈ (0, π), we can write: cot-1(x) = π/2 - y. Substituting y = tan-1(x), we get: cot-1(x) = π/2 - tan-1(x). Rearranging gives tan-1(x) + cot-1(x) = π/2.

4. Addition and Subtraction Formulas

These formulas are essential for combining inverse trigonometric functions.

• tan-1(x) + tan-1(y) = tan-1( (x+y)/(1-xy) ), if xy < 1
• tan-1(x) + tan-1(y) = π + tan-1( (x+y)/(1-xy) ), if xy > 1 and x > 0, y > 0
• tan-1(x) + tan-1(y) = -π + tan-1( (x+y)/(1-xy) ), if xy > 1 and x < 0, y < 0
• tan-1(x) + tan-1(y) = π/2, if xy = 1 and x > 0, y > 0
• tan-1(x) + tan-1(y) = -π/2, if xy = 1 and x < 0, y < 0
• tan-1(x) - tan-1(y) = tan-1( (x-y)/(1+xy) ), if xy > -1
• tan-1(x) - tan-1(y) = π + tan-1( (x-y)/(1+xy) ), if xy < -1 and x > 0, y < 0
• tan-1(x) - tan-1(y) = -π + tan-1( (x-y)/(1+xy) ), if xy < -1 and x < 0, y > 0

Let's derive the first formula: tan-1(x) + tan-1(y) = tan-1( (x+y)/(1-xy) ), if xy < 1. Let A = tan-1(x) and B = tan-1(y). Then tan(A) = x and tan(B) = y. The range of tan-1 is (-π/2, π/2). So, A ∈ (-π/2, π/2) and B ∈ (-π/2, π/2). Thus, A + B ∈ (-π, π). We know the identity tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A)tan(B)). Substituting, tan(A + B) = (x + y) / (1 - xy). If xy < 1, then 1 - xy > 0. Also, if x > 0 and y > 0, then A ∈ (0, π/2) and B ∈ (0, π/2), so A + B ∈ (0, π). If xy < 1, 1-xy > 0. If x > 0, y > 0, then x+y > 0. So (x+y)/(1-xy) > 0. If x < 0, y < 0, then x+y < 0. So (x+y)/(1-xy) < 0. If x > 0, y < 0 (or vice versa), the sign of x+y depends on their magnitudes. Consider the case where xy < 1. This implies that A+B could be in (-π/2, π/2) or outside this range but summing to a value whose tangent matches. If xy < 1, then the denominator 1 - xy is positive. If A+B is in the range (-π/2, π/2), then tan-1(tan(A+B)) = A+B. The condition xy < 1 ensures that A+B does not cross the boundaries ±π/2 where the tangent function is undefined. Specifically, if xy < 1, then A+B is not equal to π/2 or -π/2. If xy < 1, then A + B lies in the interval (-π/2, π/2). To see this, assume A + B ≥ π/2. Then B ≥ π/2 - A. Since tan is increasing in (-π/2, π/2), tan(B) ≥ tan(π/2 - A). y ≥ cot(A) = 1/x. So, xy ≥ 1. This contradicts xy < 1. Similarly, assume A + B ≤ -π/2. Then B ≤ -π/2 - A. tan(B) ≤ tan(-π/2 - A) = cot(A) = 1/x. y ≤ 1/x. So, xy ≤ 1. This doesn't strictly contradict xy < 1 if equality holds. Let's be more precise. If xy < 1, then tan(A+B) = (x+y)/(1-xy). The range of A+B is (-π, π). The range of tan-1( (x+y)/(1-xy) ) is (-π/2, π/2). If A+B falls within (-π/2, π/2), then A+B = tan-1( (x+y)/(1-xy) ). This happens when -π/2 < A+B < π/2. This condition is met when xy < 1. If xy > 1: If x > 0, y > 0, then A ∈ (0, π/2), B ∈ (0, π/2), so A+B ∈ (0, π). Since xy > 1, 1-xy < 0. If x > 0, y > 0, then x+y > 0. So (x+y)/(1-xy) < 0. This means tan(A+B) is negative. If A+B ∈ (0, π) and tan(A+B) < 0, then A+B ∈ (π/2, π). In this case, tan-1(tan(A+B)) = A+B - π. So, tan-1( (x+y)/(1-xy) ) = A+B - π. Rearranging, A+B = π + tan-1( (x+y)/(1-xy) ). This matches the second formula. If x < 0, y < 0, then A ∈ (-π/2, 0), B ∈ (-π/2, 0), so A+B ∈ (-π, 0). Since xy > 1, 1-xy < 0. If x < 0, y < 0, then x+y < 0. So (x+y)/(1-xy) > 0. This means tan(A+B) is positive. If A+B ∈ (-π, 0) and tan(A+B) > 0, then A+B ∈ (-π, -π/2). In this case, tan-1(tan(A+B)) = A+B + π. So, tan-1( (x+y)/(1-xy) ) = A+B + π. Rearranging, A+B = -π + tan-1( (x+y)/(1-xy) ). This matches the third formula.

The formulas for subtraction follow similarly.

5. Formulas for 2 tan^-1(x)

These are derived from the addition formulas by setting y = x.

• 2 tan-1(x) = tan-1( 2x/(1-x2) ), if |x| < 1
• 2 tan-1(x) = π + tan-1( 2x/(1-x2) ), if x > 1
• 2 tan-1(x) = -π + tan-1( 2x/(1-x2) ), if x < -1
• 2 tan-1(x) = π/2, if x = 1
• 2 tan-1(x) = -π/2, if x = -1

There are also alternative forms for 2 tan-1(x) in terms of sine and cosine, which are very useful in integration and other areas.

• 2 tan-1(x) = sin-1( 2x/(1+x2) ), if |x| ≤ 1
• 2 tan-1(x) = cos-1( (1-x2)/(1+x2) ), if x ≥ 0
• 2 tan-1(x) = -cos-1( (1-x2)/(1+x2) ), if x ≤ 0 (This is incorrect, should be 2π + cos-1(...) or similar depending on range)
• Let's re-derive the cosine form. Let θ = tan-1(x). Then tan(θ) = x. We want to express cos(2θ) in terms of x. We know cos(2θ) = (1 - tan2(θ)) / (1 + tan2(θ)). So, cos(2θ) = (1 - x2) / (1 + x2). This means 2θ = cos-1( (1-x2)/(1+x2) ), provided that 2θ is in the range [0, π]. If x ≥ 0, then θ = tan-1(x) ∈ [0, π/2). So 2θ ∈ [0, π). This range is valid for cos-1. Thus, 2 tan-1(x) = cos-1( (1-x2)/(1+x2) ) for x ≥ 0. If x < 0, then θ = tan-1(x) ∈ (-π/2, 0). So 2θ ∈ (-π, 0). The value cos-1( (1-x2)/(1+x2) ) will be in (0, π] because (1-x2)/(1+x2) ranges from -1 (exclusive) to 1 (exclusive). Specifically, if x < 0, let x = -a where a > 0. Then (1 - (-a)2) / (1 + (-a)2) = (1 - a2) / (1 + a2). So, cos(2θ) = (1 - a2) / (1 + a2). Let Φ = cos-1( (1-a2)/(1+a2) ). Since a > 0, Φ ∈ [0, π). We have 2θ ∈ (-π, 0). The identity cos(α) = cos(β) implies α = ±β + 2nπ. Here, cos(2θ) = cos(Φ). So, 2θ = ±Φ + 2nπ. Since 2θ ∈ (-π, 0) and Φ ∈ [0, π). If we take the positive sign, 2θ = Φ + 2nπ. For n=0, 2θ = Φ, which is impossible as LHS is negative and RHS is non-negative. For n=-1, 2θ = Φ - 2π. This is possible. If we take the negative sign, 2θ = -Φ + 2nπ. For n=0, 2θ = -Φ. This is possible if Φ ≠ 0. Let's check the range of (1-x2)/(1+x2). As x goes from -∞ to 0, x2 goes from ∞ to 0. So 1-x2 goes from -∞ to 1. 1+x2 goes from ∞ to 1. The ratio goes from -1 (exclusive) to 1 (exclusive). So cos-1( (1-x2)/(1+x2) ) is in (0, π). Thus, for x < 0, 2 tan-1(x) ∈ (-π, 0) and cos-1( (1-x2)/(1+x2) ) ∈ (0, π). The relation is 2 tan-1(x) = - cos-1( (1-x2)/(1+x2) ). So, the correct forms are: 2 tan-1(x) = cos-1( (1-x2)/(1+x2) ), for x ≥ 0 2 tan-1(x) = -cos-1( (1-x2)/(1+x2) ), for x < 0

6. Formulas for 3 tan^-1(x)

These can be derived using 3θ = 2θ + θ.

• 3 tan-1(x) = tan-1( (3x-x3)/(1-3x2) ), provided that 3x-x3 and 1-3x2 have the same sign and |3x-1| < 1. More general forms exist with ±π adjustments.

7. Formulas involving sin^-1(x) and cos^-1(x)

These are often derived by converting to tan^-1 or by using angle addition formulas for sine and cosine.

• sin-1(x) + sin-1(y) = sin-1( x√(1-y2) + y√(1-x2) ), if x2 + y2 ≤ 1 or if x, y > 0 and x2 + y2 ≤ 1. Adjustments of ±π are needed for other cases.
• cos-1(x) + cos-1(y) = cos-1( xy - √(1-x2)√(1-y2) ), if x + y ≥ 0. Adjustments of ±2π are needed otherwise.

Solving Equations using Inverse Trigonometric Functions

Inverse trigonometric functions are frequently used to solve trigonometric equations. The process typically involves:

1. Isolating the trigonometric function.
2. Applying the corresponding inverse trigonometric function to both sides.
3. Using the properties of inverse trigonometric functions to simplify and solve for the variable.

Example: Solve for x in the equation 2 sin(x) = 1.

1. Isolate sin(x): sin(x) = 1/2.
2. Apply inverse sine: x = sin-1(1/2).
3. Evaluate: We know that sin(π/6) = 1/2 and π/6 is in the principal range [-π/2, π/2]. So, the principal solution is x = π/6.
4. General solution: Since the sine function is periodic, the general solution is x = nπ + (-1)n(π/6), where n is an integer.

Example: Solve tan-1(2x) + tan-1(3x) = π/4.

1. Use the addition formula for tan-1. Here, x becomes 2x and y becomes 3x.
2. We need to check the condition for the formula. If (2x)(3x) < 1, i.e., 6x2 < 1, then: tan-1( (2x+3x)/(1 - (2x)(3x)) ) = π/4 tan-1( 5x / (1 - 6x2) ) = π/4
3. Apply tangent to both sides: 5x / (1 - 6x2) = tan(π/4) = 1 5x = 1 - 6x2 6x2 + 5x - 1 = 0
4. Solve the quadratic equation: (6x - 1)(x + 1) = 0 So, x = 1/6 or x = -1.
5. Check the condition 6x2 < 1. If x = 1/6, then 6(1/6)2 = 6(1/36) = 1/6. Since 1/6 < 1, this solution is valid under this condition. If x = -1, then 6(-1)2 = 6. Since 6 > 1, this solution is not valid under the condition xy < 1.
6. Consider the case where xy > 1. Here, 6x2 > 1. If x > 0 (and thus y > 0, since x = -1 is not positive), the formula is: π + tan-1( 5x / (1 - 6x2) ) = π/4 tan-1( 5x / (1 - 6x2) ) = π/4 - π = -3π/4 5x / (1 - 6x2) = tan(-3π/4) = 1 This leads to the same quadratic equation 6x2 + 5x - 1 = 0, with solutions x = 1/6 and x = -1. We need x > 0 for this case, so x = 1/6 is considered. But we assumed 6x2 > 1, and for x=1/6, 6x2 = 1/6 < 1. So this case doesn't yield a solution. If x < 0 (and thus y < 0), the formula is: -π + tan-1( 5x / (1 - 6x2) ) = π/4 tan-1( 5x / (1 - 6x2) ) = π/4 + π = 5π/4 This is impossible, as the range of tan-1 is (-π/2, π/2).
7. Let's re-evaluate the solution x = -1. If x = -1, then 2x = -2 and 3x = -3. tan-1(-2) + tan-1(-3). Both arguments are negative. xy = (-2)(-3) = 6 > 1. The formula is -π + tan-1( ( -2 + -3 ) / (1 - (-2)(-3)) ) = -π + tan-1( -5 / (1 - 6) ) = -π + tan-1( -5 / -5 ) = -π + tan-1(1) = -π + π/4 = -3π/4. This is not equal to π/4. So x = -1 is not a solution.
8. Therefore, the only solution is x = 1/6.

Graphical Representation

The graphs of inverse trigonometric functions are reflections of the original trigonometric functions across the line y = x, restricted to their principal value ranges.

• y = sin-1(x): Starts at (-1, -π/2), passes through (0, 0), and ends at (1, π/2). It is an increasing function.
• y = cos-1(x): Starts at (-1, π), passes through (0, π/2), and ends at (1, 0). It is a decreasing function.
• y = tan-1(x): Has horizontal asymptotes at y = π/2 and y = -π/2. It passes through (0, 0) and increases throughout its domain.

Understanding these graphs helps in visualizing the domain, range, and behavior of these functions.