Basic Concepts and Principal Values

Basic Concepts and Principal Values

Inverse trigonometric functions are used to find the angle when the value of a trigonometric ratio is known. For example, if sin θ = 1/2, then we can use sin⁻¹(1/2) to find an angle whose sine is 1/2. लेकिन यहाँ एक important point है: एक ही trigonometric ratio का value कई अलग-अलग angles पर हो सकता है, इसलिए inverse trigonometric functions के लिए एक निश्चित interval choose करना पड़ता है। इसी कारण principal value का concept आता है।

1. Inverse Trigonometric Functions

For a function to have an inverse, it must be one-one on its domain. Since the usual trigonometric functions are not one-one over their entire domains, their domains are restricted to suitable intervals so that inverse functions can be defined.

For example, sin x is not one-one on the whole real line because different values of x can have the same sine. लेकिन यदि sin x को एक suitable interval पर restrict कर दिया जाए, तो वह one-one बन जाता है और उसका inverse define किया जा सकता है।

The inverse trigonometric functions are written as:

Trigonometric Function	Inverse Function
sin x	sin−1 x
cos x	cos−1 x
tan x	tan−1 x
cot x	cot−1 x
sec x	sec−1 x
cosec x	cosec−1 x

2. Meaning of sin⁻¹ x

If:

y = sin⁻¹x

then:

sin y = x

and y is taken from the principal value interval of sin⁻¹x.

उदाहरण के लिए:

sin⁻¹(1/2) = π/6

क्योंकि sin(π/6) = 1/2 और π/6 principal value interval में आता है।

3. Principal Value

A trigonometric equation may have infinitely many solutions, but an inverse trigonometric function gives a unique value by restricting the angle to a particular interval. This selected value is called the principal value.

यानी किसी trigonometric ratio के लिए कई angles possible हो सकते हैं, लेकिन inverse function उनमें से एक निश्चित angle को answer के रूप में चुनता है। वही principal value कहलाती है।

4. Principal Value of sin⁻¹ x

For sin⁻¹x, the principal value lies in:

−π/2 ≤ sin⁻¹x ≤ π/2

Therefore:

Range of sin⁻¹x = [−π/2, π/2]

इस interval में sine one-one रहता है, इसलिए हर allowed value के लिए principal value uniquely निर्धारित होती है।

Examples

sin⁻¹(0) = 0

sin⁻¹(1/2) = π/6

sin⁻¹(√3/2) = π/3

sin⁻¹(−1/2) = −π/6

5. Principal Value of cos⁻¹ x

For cos⁻¹x, the principal value lies in:

0 ≤ cos⁻¹x ≤ π

Therefore:

Range of cos⁻¹x = [0, π]

इस interval में cosine one-one होता है, इसलिए inverse cosine का answer इसी interval में लिया जाता है।

Examples

cos⁻¹(1) = 0

cos⁻¹(1/2) = π/3

cos⁻¹(0) = π/2

cos⁻¹(−1) = π

6. Principal Value of tan⁻¹ x

For tan⁻¹x, the principal value lies in:

−π/2 < tan⁻¹x < π/2

Therefore:

Range of tan⁻¹x = (−π/2, π/2)

ध्यान दें कि यहाँ −π/2 और π/2 included नहीं हैं क्योंकि tan x इन angles पर defined नहीं है।

Examples

tan⁻¹(0) = 0

tan⁻¹(1) = π/4

tan⁻¹(−1) = −π/4

7. Principal Value of cot⁻¹ x

For cot⁻¹x, the principal value is taken in:

0 < cot⁻¹x < π

इसका मतलब है कि principal value 0 और π के बीच रहती है।

Examples

cot⁻¹(1) = π/4

cot⁻¹(0) = π/2

cot⁻¹(−1) = 3π/4

8. Principal Value of sec⁻¹ x

For sec⁻¹x, the principal value is taken in:

[0, π] \ {π/2}

और sec⁻¹x केवल उन x values के लिए defined है जिनके लिए:

x ∈ (−∞, −1] ∪ [1, ∞)

क्योंकि sec θ का value −1 और 1 के बीच नहीं हो सकता।

Examples

sec⁻¹(1) = 0

sec⁻¹(2) = π/3

sec⁻¹(−1) = π

9. Principal Value of cosec⁻¹ x

For cosec⁻¹x, the principal value is taken in:

[−π/2, π/2] \ {0}

और cosec⁻¹x के लिए:

x ∈ (−∞, −1] ∪ [1, ∞)

क्योंकि cosec θ का absolute value 1 से छोटा नहीं हो सकता।

Examples

cosec⁻¹(1) = π/2

cosec⁻¹(−1) = −π/2

cosec⁻¹(2) = π/6

10. Principal Value Table

Function	Domain	Principal Value Range
sin−1x	[−1, 1]	[−π/2, π/2]
cos−1x	[−1, 1]	[0, π]
tan−1x	R	(−π/2, π/2)
cot−1x	R	(0, π)
sec−1x	(−∞,−1] ∪ [1,∞)	[0,π] excluding π/2
cosec−1x	(−∞,−1] ∪ [1,∞)	[−π/2,π/2] excluding 0

11. Basic Principal Values

Expression	Principal Value
sin−1(0)	0
sin−1(1/2)	π/6
sin−1(√3/2)	π/3
sin−1(1)	π/2
cos−1(1)	0
cos−1(√3/2)	π/6
cos−1(1/2)	π/3
cos−1(0)	π/2
tan−1(0)	0
tan−1(1)	π/4
tan−1(√3)	π/3
cot−1(1)	π/4
cot−1(√3)	π/6

12. How to Find a Principal Value

When finding an inverse trigonometric value, first identify the corresponding standard angle and then check whether that angle lies in the principal value range.

अगर standard angle principal interval में है, तो वही answer होगा। यदि वह interval से बाहर है, तो ऐसा दूसरा angle चुनना होगा जो same trigonometric value देता हो और principal interval के अंदर हो।

Example 1

Find:

sin⁻¹(−1/2)

We know:

sin(−π/6) = −1/2

Since −π/6 lies in [−π/2, π/2], therefore:

sin⁻¹(−1/2) = −π/6

Example 2

Find:

cos⁻¹(−1/2)

We know:

cos(2π/3) = −1/2

Since 2π/3 lies in [0,π], therefore:

cos⁻¹(−1/2) = 2π/3

Example 3

Find:

tan⁻¹(−1)

We know:

tan(−π/4) = −1

Since −π/4 lies in (−π/2,π/2), therefore:

tan⁻¹(−1) = −π/4

13. Important Identities

For x ∈ [−1,1]:

sin(sin⁻¹x) = x

For x ∈ [−1,1]:

cos(cos⁻¹x) = x

For x ∈ R:

tan(tan⁻¹x) = x

Similarly:

cot(cot⁻¹x) = x

sec(sec⁻¹x) = x

cosec(cosec⁻¹x) = x

14. Example: sin⁻¹(sin x)

Consider:

sin⁻¹(sin 5π/6)

Since:

sin(5π/6) = 1/2

we get:

sin⁻¹(1/2) = π/6

Therefore:

sin⁻¹(sin 5π/6) = π/6

यहाँ answer 5π/6 नहीं आया क्योंकि sin⁻¹x का principal value केवल [−π/2,π/2] में लिया जाता है।

15. Example: cos⁻¹(cos x)

Consider:

cos⁻¹(cos 4π/3)

We know:

cos(4π/3) = −1/2

The angle in the principal range [0,π] having cosine −1/2 is 2π/3.

Therefore:

cos⁻¹(cos 4π/3) = 2π/3

16. Key Points to Remember

• Inverse trigonometric functions give angles corresponding to given trigonometric ratios.
• Principal value makes the inverse trigonometric function single-valued.
• sin⁻¹x has principal values in [−π/2,π/2].
• cos⁻¹x has principal values in [0,π].
• tan⁻¹x has principal values in (−π/2,π/2).
• cot⁻¹x has principal values in (0,π).
• sec⁻¹x has principal values in [0,π] excluding π/2.
• cosec⁻¹x has principal values in [−π/2,π/2] excluding 0.
• Always check the principal value range before writing the final answer.

17. Practice Questions

1. Find sin⁻¹(1/2).
2. Find sin⁻¹(−√3/2).
3. Find cos⁻¹(1/2).
4. Find cos⁻¹(−1/2).
5. Find tan⁻¹(1).
6. Find tan⁻¹(−√3).
7. Find cot⁻¹(1).
8. Find cot⁻¹(−1).
9. Find sec⁻¹(2).
10. Find cosec⁻¹(2).
11. Evaluate sin⁻¹(sin 5π/6).
12. Evaluate cos⁻¹(cos 4π/3).
13. Explain why sin⁻¹(sin x) is not always equal to x.
14. State the principal value ranges of all six inverse trigonometric functions.

18. Multiple Choice Questions

1. The principal value range of sin⁻¹x is:

1. [0,π]
2. [−π/2,π/2]
3. (−π/2,π/2)
4. (0,π)

Answer: (b)

2. The principal value of cos⁻¹(−1/2) is:

1. π/3
2. −π/3
3. 2π/3
4. π/6

Answer: (c)

3. The value of tan⁻¹(1) is:

1. π/6
2. π/4
3. π/3
4. π/2

Answer: (b)

4. The domain of sin⁻¹x is:

1. R
2. [0,∞)
3. [−1,1]
4. (−∞,∞)

Answer: (c)

19. Quick Revision

Inverse Function	Principal Value Range
sin−1x	[−π/2, π/2]
cos−1x	[0, π]
tan−1x	(−π/2, π/2)
cot−1x	(0, π)
sec−1x	[0,π] excluding π/2
cosec−1x	[−π/2,π/2] excluding 0

Remember: Inverse Trigonometric Function → Given ratio से angle find करना → Principal Value Range check करना → Final answer लिखना.