What is the value of \({\sin ^{ - 1}}\frac{4}{5} + {\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2}?\)
0
The question asks for the value of the expression \({\sin ^{ - 1}}\frac{4}{5} + {\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2}\). To solve this, we need to use properties of inverse trigonometric functions.
Inverse trigonometric functions, also known as arc functions, are the inverse functions of the trigonometric functions. They provide the angle whose trigonometric function is equal to a given value. For example, \({\sin ^{ - 1}}x\) gives the angle \(\theta\) such that \(\sin \theta = x\).
We can use specific identities to simplify expressions involving inverse trigonometric functions.
The expression is \({\sin ^{ - 1}}\frac{4}{5} + {\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2}\). Let's analyze each term:
The first term is \({\sin ^{ - 1}}\frac{4}{5}\). This is the angle whose sine is \(\frac{4}{5}\).
The second term is \({\sec ^{ - 1}}\frac{5}{4}\). This is the angle whose secant is \(\frac{5}{4}\).
The third term is \(-\frac{\pi}{2}\).
We know the identity relating inverse secant and inverse cosine:
\({\sec ^{ - 1}}x = {\cos ^{ - 1}}\frac{1}{x}\), provided that \(|x| \ge 1\).
In our expression, the second term is \({\sec ^{ - 1}}\frac{5}{4}\). Here, \(x = \frac{5}{4}\). Since \(|\frac{5}{4}| = \frac{5}{4} \ge 1\), we can apply the identity:
\({\sec ^{ - 1}}\frac{5}{4} = {\cos ^{ - 1}}\frac{1}{{5/4}} = {\cos ^{ - 1}}\frac{4}{5}\).
Now, substitute this back into the original expression:
\({\sin ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{4}{5} - \frac{\pi }{2}\).
We also know a fundamental identity involving inverse sine and inverse cosine:
\({\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}\), provided that \(x \in [-1, 1]\).
In our simplified expression, we have \({\sin ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{4}{5}\). Here, \(x = \frac{4}{5}\). Since \(\frac{4}{5}\) is in the interval \([-1, 1]\), we can apply this identity:
\({\sin ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{4}{5} = \frac{\pi }{2}\).
Substitute this value back into the expression:
\(\frac{\pi }{2} - \frac{\pi }{2}\).
Now, perform the final subtraction:
\(\frac{\pi }{2} - \frac{\pi }{2} = 0\).
Therefore, the value of the given expression is \(0\).
| Step | Expression | Identity Used |
|---|---|---|
| 1 | \({\sin ^{ - 1}}\frac{4}{5} + {\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2}\) | Original expression |
| 2 | \({\sec ^{ - 1}}\frac{5}{4} = {\cos ^{ - 1}}\frac{4}{5}\) | \({\sec ^{ - 1}}x = {\cos ^{ - 1}}\frac{1}{x}\) for \(|x| \ge 1\) |
| 3 | \({\sin ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{4}{5} - \frac{\pi }{2}\) | Substitute result from Step 2 into Step 1 |
| 4 | \({\sin ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{4}{5} = \frac{\pi }{2}\) | \({\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}\) for \(x \in [-1, 1]\) |
| 5 | \(\frac{\pi }{2} - \frac{\pi }{2}\) | Substitute result from Step 4 into Step 3 |
| 6 | \(0\) | Final calculation |
| Identity | Condition |
|---|---|
| \({\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2}\) | \(x \in [-1, 1]\) |
| \({\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}\) | \(x \in (-\infty, \infty)\) |
| \({\sec ^{ - 1}}x + {\csc ^{ - 1}}x = \frac{\pi }{2}\) | \(|x| \ge 1\) |
| \({\sec ^{ - 1}}x = {\cos ^{ - 1}}\frac{1}{x}\) | \(|x| \ge 1\) |
| \({\csc ^{ - 1}}x = {\sin ^{ - 1}}\frac{1}{x}\) | \(|x| \ge 1\) |
| \({\cot ^{ - 1}}x = {\tan ^{ - 1}}\frac{1}{x}\) | \(x > 0\) |
| \({\cot ^{ - 1}}x = \pi + {\tan ^{ - 1}}\frac{1}{x}\) | \(x < 0\) |
When working with inverse trigonometric functions, it's important to remember their specific domains and ranges to ensure the identities are applied correctly. The principal value branches are usually used.
\(y = {\sin ^{ - 1}}x\): Domain \([-1, 1]\), Range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
\(y = {\cos ^{ - 1}}x\): Domain \([-1, 1]\), Range \([0, \pi]\).
\(y = {\tan ^{ - 1}}x\): Domain \((-\infty, \infty)\), Range \((-\frac{\pi}{2}, \frac{\pi}{2})\).
\(y = {\cot ^{ - 1}}x\): Domain \((-\infty, \infty)\), Range \((0, \pi)\).
\(y = {\sec ^{ - 1}}x\): Domain \((-\infty, -1] \cup [1, \infty)\), Range \([0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]\).
\(y = {\csc ^{ - 1}}x\): Domain \((-\infty, -1] \cup [1, \infty)\), Range \([-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]\).
In this problem, \(\frac{4}{5}\) is in the domain of \({\sin ^{ - 1}}x\) and \({\cos ^{ - 1}}x\), and \(\frac{5}{4}\) is in the domain of \({\sec ^{ - 1}}x\), satisfying the conditions for the identities used.
What is 2 cot \(\left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\) equal to ?
The equation \({\tan ^{ - 1}}\left( {1 + {\rm{x}}} \right) + {\tan ^{ - 1}}\left( {1 - {\rm{x}}} \right) = \frac{{\rm{\pi }}}{2}\) is satisfied by
The equation \(sin^{-1}x-cos^{-1}x=\frac{\pi}{6}\) has
What is \(\tan \left\{ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right\}\) equal to?
If \({\sin ^{ - 1}}\frac{{2p}}{{1 + p2}} - {\cos ^{ - 1}}\frac{{1 - {q^2}}}{{1 + {q^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\) , then what is x equal to?
Consider the following values of x:
1) 8
2) -4
3) \(\frac 16\)
4) \(- \frac{1}{4}\)
Which of the above values of x is/are the solution(s) of the equation
\({\tan ^{ - 1}}\left( {2x} \right) + {\tan ^{ - 1}}\left( {3x} \right) = \frac{\pi }{4}?{\rm{\;}}\)
What is \(\tan ^{- 1}\left( {\frac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{3}{5}} \right)\) equal to?
Let the equation sec x.cosec x = p have a solution, where p is a positive real number. What should be the smallest value of p?
For what value of θ, where 0 < θ < \(\frac{\pi}{2}\) , does sin θ + sin θ cos θ maximum value?
What is 2 cot \(\left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\) equal to ?
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