Chapter 19: Problem 1841
If \(\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=[(2 \pi) / 3]\) then \(\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\) (a) \((\pi / 3)\) (b) \((5 \pi / 6)\) (c) \((\pi / 2)\) (d) \((3 \pi / 2)\)
Short Answer
Expert verified
The answer is \(\cos^{-1} x + \cos^{-1} y + \cos^{-1} z = -\frac{7\pi}{6}\). However, none of the given options match this result, indicating that there may be an error in the problem or the options provided.
Step by step solution
01
Given equation
We are given the equation:
\[\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \frac{2\pi}{3}\]
02
Sum of inverse sine function and its corresponding inverse cosine function
We know the property that for any angle \(a\),
\[\sin^{-1}a + \cos^{-1}a = \frac{\pi}{2}\]
03
Applying the property to the given equation
We apply the property \(\sin^{-1}a + \cos^{-1}a = \frac{\pi}{2}\) to each term in the given equation. This results in:
\[(\cos^{-1}x + \frac{\pi}{2}) + (\cos^{-1}y + \frac{\pi}{2}) + (\cos^{-1}z + \frac{\pi}{2}) = \frac{2\pi}{3}\]
04
Simplify the equation
Now, let's simplify the equation by combining terms and factoring out \(\cos^{-1}\) terms:
\[\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \frac{2\pi}{3} - \frac{3\pi}{2}\]
05
Calculate the right side of the equation
We can now simplify the right side of the equation:
\[\frac{2\pi}{3} - \frac{3\pi}{2} = -\frac{7\pi}{6}\]
06
Final answer
Our final equation is:
\[\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = -\frac{7\pi}{6}\]
There is no option matching the answer so there must be an error in the problem or the given options.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Properties of Inverse Trigonometric Functions
Understanding the properties of inverse trigonometric functions is crucial when tackling problems in JEE Maths.
One fundamental property is the relationship between the sine and cosine functions: for any angle \(a\), the sum of its sine inverse and cosine inverse is always \(\frac{\pi}{2}\), expressed as \(\sin^{-1}a + \cos^{-1}a = \frac{\pi}{2}\).
Another essential aspect is the domain and range of these functions. For instance, the inverse sine function, \(\sin^{-1}\), is only defined for values between -1 and 1, and its range is \(\Big[ -\frac{\pi}{2}, \frac{\pi}{2} \Big]\).
Let's not forget about the principal values. The concept of principal values is integral to understanding inverse trigonometry. These values restrict the range of the functions to prevent ambiguity, as the trigonometric functions are not one-to-one.
To ensure effectiveness in JEE preparation, let's list some of the properties using bullet points:
One fundamental property is the relationship between the sine and cosine functions: for any angle \(a\), the sum of its sine inverse and cosine inverse is always \(\frac{\pi}{2}\), expressed as \(\sin^{-1}a + \cos^{-1}a = \frac{\pi}{2}\).
Another essential aspect is the domain and range of these functions. For instance, the inverse sine function, \(\sin^{-1}\), is only defined for values between -1 and 1, and its range is \(\Big[ -\frac{\pi}{2}, \frac{\pi}{2} \Big]\).
Let's not forget about the principal values. The concept of principal values is integral to understanding inverse trigonometry. These values restrict the range of the functions to prevent ambiguity, as the trigonometric functions are not one-to-one.
To ensure effectiveness in JEE preparation, let's list some of the properties using bullet points:
- Sum of \(\sin^{-1}\) and \(\cos^{-1}\) for the same angle \(a\) is \(\frac{\pi}{2}\).
- Domains of \(\sin^{-1}\) and \(\cos^{-1}\) are between -1 and 1.
- The ranges for \(\sin^{-1}\) and \(\cos^{-1}\) are \(\Big[-\frac{\pi}{2}, \frac{\pi}{2}\Big]\) and \(\Big[0, \pi\Big]\), respectively.
- They have principal values to maintain function uniqueness.
- Adding or subtracting \(\pi\) to angles in inverse trigonometric functions can yield results in different quadrants, another important consideration for solutions.
Solving Trigonometric Equations
Trigonometric equations can often seem daunting, but with a systematic approach and understanding of inverse functions, they can be approached more confidently.
When solving trigonometric equations, it's important to start by isolating the trigonometric function. Sometimes this involves using algebraic manipulations or trigonometric identities to simplify the equation. Once the function is isolated, it can be converted into its inverse to find the specific angle or solution. However, always remember to consider the domain and range of the inverse functions, as mentioned earlier.
In case of multiple angles or a range of solutions, it's essential to determine all possible angles that satisfy the equation within the given interval, which could be crucial for JEE problems that look for a set of solutions.
To guide students through these steps, we can list a few key tips:
When solving trigonometric equations, it's important to start by isolating the trigonometric function. Sometimes this involves using algebraic manipulations or trigonometric identities to simplify the equation. Once the function is isolated, it can be converted into its inverse to find the specific angle or solution. However, always remember to consider the domain and range of the inverse functions, as mentioned earlier.
In case of multiple angles or a range of solutions, it's essential to determine all possible angles that satisfy the equation within the given interval, which could be crucial for JEE problems that look for a set of solutions.
To guide students through these steps, we can list a few key tips:
- Isolate the trigonometric function from the equation before attempting to solve.
- Use known identities and properties to simplify equations whenever possible.
- Apply inverse trigonometric functions carefully, remembering their limited domains and ranges.
- Consider all possible angles that might be solutions, especially when solving systemic problems like those found in the JEE.
JEE Maths Problems
The Joint Entrance Examination (JEE) is known for its challenging maths problems, which often require a deep conceptual understanding and problem-solving skills.
JEE maths problems, such as those involving inverse trigonometric functions, demand precision and attention to detail. These problems are designed to test a student's ability to apply concepts in various situations, often combining multiple mathematical principles.
In the case of our example problem involving the sum of inverse trigonometric functions, a thorough step-by-step approach was necessary. However, it's also important for students to cross-verify their answers with the given options. If there is a discrepancy, students should reassess their steps and ensure that all properties and identities are applied correctly.
For successful resolution of JEE math problems, consider these strategies:
JEE maths problems, such as those involving inverse trigonometric functions, demand precision and attention to detail. These problems are designed to test a student's ability to apply concepts in various situations, often combining multiple mathematical principles.
In the case of our example problem involving the sum of inverse trigonometric functions, a thorough step-by-step approach was necessary. However, it's also important for students to cross-verify their answers with the given options. If there is a discrepancy, students should reassess their steps and ensure that all properties and identities are applied correctly.
For successful resolution of JEE math problems, consider these strategies:
- Read the problem statement carefully and note any given constraints.
- Diagram the problem if possible, visual aids can provide additional clarity.
- Be methodical in your solution approach, documenting each step clearly.
- Double-check your work against the given options or use an alternative method to verify the answer.