The equation \(sin^{-1}x-cos^{-1}x=\frac{\pi}{6}\) has
unique solution
Solving Inverse Trigonometric Equations
We are asked to find the number of solutions for the equation \(sin^{-1}x - cos^{-1}x = \frac{\pi}{6}\).
To solve this equation involving inverse trigonometric functions, we can use a known identity. The fundamental identity relating \(sin^{-1}x\) and \(cos^{-1}x\) is:
\(sin^{-1}x + cos^{-1}x = \frac{\pi}{2}\)
This identity is valid for all values of \(x\) in the domain of both functions, which is \([-1, 1]\).
We now have a system of two linear equations with \(sin^{-1}x\) and \(cos^{-1}x\) as the variables:
- \(sin^{-1}x - cos^{-1}x = \frac{\pi}{6}\) (Given equation)
- \(sin^{-1}x + cos^{-1}x = \frac{\pi}{2}\) (Identity)
We can solve this system by adding the two equations together:
\((sin^{-1}x - cos^{-1}x) + (sin^{-1}x + cos^{-1}x) = \frac{\pi}{6} + \frac{\pi}{2}\)
Simplifying the left side:
\(2 sin^{-1}x = \frac{\pi}{6} + \frac{3\pi}{6}\)
\(2 sin^{-1}x = \frac{4\pi}{6}\)
\(2 sin^{-1}x = \frac{2\pi}{3}\)
Now, divide both sides by 2 to isolate \(sin^{-1}x\):
\(sin^{-1}x = \frac{1}{2} \times \frac{2\pi}{3}\)
\(sin^{-1}x = \frac{\pi}{3}\)
To find the value of \(x\), we take the sine of both sides:
\(x = sin(\frac{\pi}{3})\)
We know that \(sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\). So,
\(x = \frac{\sqrt{3}}{2}\)
Now, we must check if this value of \(x\) is valid. The domain of \(sin^{-1}x\) and \(cos^{-1}x\) is \([-1, 1]\). The value \(\frac{\sqrt{3}}{2}\) is approximately \(0.866\), which is within the interval \([-1, 1]\).
To confirm, let's substitute \(x = \frac{\sqrt{3}}{2}\) back into the original equation:
\(sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}\) (since \(sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\) and \(\frac{\pi}{3}\) is in the range of \(sin^{-1}x\), \([-\frac{\pi}{2}, \frac{\pi}{2}]\))
\(cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{6}\) (since \(cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}\) and \(\frac{\pi}{6}\) is in the range of \(cos^{-1}x\), \([0, \pi]\))
Now, check the original equation:
\(sin^{-1}x - cos^{-1}x = \frac{\pi}{3} - \frac{\pi}{6}\)
To subtract, find a common denominator:
\(\frac{2\pi}{6} - \frac{\pi}{6} = \frac{\pi}{6}\)
The value \(x = \frac{\sqrt{3}}{2}\) satisfies the original equation.
Since we found exactly one valid value for \(x\), the equation has a unique solution.
Analysis of Equation Solutions
We found one specific value of \(x\) that satisfies the given equation \(sin^{-1}x - cos^{-1}x = \frac{\pi}{6}\). This means:
- There is a solution.
- The solution is \(x = \frac{\sqrt{3}}{2}\).
- There are no other possible values for \(x\) that satisfy the system of equations derived from the identity, and this solution is within the domain.
Therefore, the equation has a unique solution.
Revision Table: Inverse Trigonometric Functions
| Function | Domain | Range | Key Identity |
|---|---|---|---|
| \(sin^{-1}x\) | \([-1, 1]\) | \([-\frac{\pi}{2}, \frac{\pi}{2}]\) | \(sin^{-1}x + cos^{-1}x = \frac{\pi}{2}\) |
| \(cos^{-1}x\) | \([-1, 1]\) | \([0, \pi]\) | \(sin^{-1}x + cos^{-1}x = \frac{\pi}{2}\) |
Additional Information on Inverse Trigonometric Equations
Inverse trigonometric functions, also known as arc functions, are the inverse functions of the trigonometric functions. They are used to find the angle when the value of the trigonometric ratio is given. For example, \(sin^{-1}y = \theta\) means \(sin(\theta) = y\).
When solving equations involving inverse trigonometric functions, it is crucial to:
- Understand the domain and range of each inverse function. The domain restricts the possible values of the input variable (like \(x\)). The range restricts the possible output angle values.
- Use appropriate identities to simplify the equation. The identity \(sin^{-1}x + cos^{-1}x = \frac{\pi}{2}\) is very common in such problems. Other useful identities involve \(tan^{-1}x\), \(cot^{-1}x\), \(sec^{-1}x\), and \(csc^{-1}x\).
- Always check the obtained solution(s) to ensure they are within the domain of the original inverse trigonometric functions in the equation. A solution might be mathematically correct from solving a derived equation but invalid if it falls outside the domain of the original functions.
- Be aware that inverse trigonometric functions are typically defined with restricted ranges to make them one-to-one and thus invertible. For example, \(sin^{-1}x\) only gives angles between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
In this specific problem, the use of the identity allowed us to convert a single equation with two different inverse functions into a system that could be solved for each inverse function separately, ultimately leading to the value of \(x\).
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The principal value of sin−1\(\frac{1}{\sqrt{2}}\) is equal to which of the following?
The imaginary part of log sin (x + iy) is:
The value of \({\tan ^{ - 1}}\left( {\frac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{3}} \right)\) is
The function \(f(x) = \sqrt {\cos (\sin x)} + {\sin ^{ - 1}}\left( {\frac{{1 + {x^2}}}{{2x}}} \right)\) is defined for
