What is \(1+\sin ^2\left(\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\) equal to ?
Evaluating Trigonometric Expressions with Inverse Functions
The question asks us to find the value of the expression \(1+\sin ^2\left(\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\).
This expression involves both a trigonometric function (\(\sin^2\)) and an inverse trigonometric function (\(\cos^{-1}\)). To evaluate this, we can use a substitution to simplify the inverse trigonometric part.
Step-by-Step Evaluation of \(1+\sin ^2\left(\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\)
- Let \(\theta = \cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\).
By the definition of the inverse cosine function, this means that \(\cos \theta = \frac{3}{\sqrt{17}}\). Here, \(\theta\) represents an angle whose cosine is \(\frac{3}{\sqrt{17}}\).
- Substitute \(\theta\) into the original expression.
The expression \(1+\sin ^2\left(\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\) becomes \(1+\sin^2(\theta)\).
- Use a trigonometric identity to relate \(\sin^2 \theta\) and \(\cos^2 \theta\).
We know the fundamental trigonometric identity: \(\sin^2 \theta + \cos^2 \theta = 1\).
From this identity, we can express \(\sin^2 \theta\) as \(\sin^2 \theta = 1 - \cos^2 \theta\).
- Substitute the value of \(\cos \theta\) into the identity.
We found that \(\cos \theta = \frac{3}{\sqrt{17}}\). Therefore, \(\cos^2 \theta = \left(\frac{3}{\sqrt{17}}\right)^2\).
\(\cos^2 \theta = \frac{3^2}{(\sqrt{17})^2} = \frac{9}{17}\).
- Calculate \(\sin^2 \theta\).
\(\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{9}{17}\).
To subtract, find a common denominator:
\(\sin^2 \theta = \frac{17}{17} - \frac{9}{17} = \frac{17-9}{17} = \frac{8}{17}\).
- Substitute the value of \(\sin^2 \theta\) back into the simplified expression \(1+\sin^2(\theta)\).
The expression is \(1+\sin^2 \theta\).
Substitute \(\sin^2 \theta = \frac{8}{17}\):
\(1 + \frac{8}{17}\).
- Calculate the final value.
\(1 + \frac{8}{17} = \frac{17}{17} + \frac{8}{17} = \frac{17+8}{17} = \frac{25}{17}\).
Thus, the value of \(1+\sin ^2\left(\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\) is \(\frac{25}{17}\).
Verification of the Result
We can summarise the calculation steps:
| Expression Part | Calculation | Value |
|---|---|---|
| \(\cos^{-1}\left(\frac{3}{\sqrt{17}}\right)\) | Let \(\theta = \cos^{-1}\left(\frac{3}{\sqrt{17}}\right)\) | \(\cos \theta = \frac{3}{\sqrt{17}}\) |
| \(\sin^2\left(\cos^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\) | \(\sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(\frac{3}{\sqrt{17}}\right)^2\) | \(1 - \frac{9}{17} = \frac{8}{17}\) |
| \(1+\sin^2\left(\cos^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\) | \(1 + \sin^2 \theta = 1 + \frac{8}{17}\) | \(\frac{25}{17}\) |
The final calculated value matches one of the given options.
Revision Table: Key Trigonometry Concepts
| Concept | Description | Formula/Identity |
|---|---|---|
| Inverse Cosine Function (\(\cos^{-1}x\)) | Returns the angle \(\theta\) such that \(\cos \theta = x\). The range is typically \(0 \le \theta \le \pi\). | If \(\cos \theta = x\), then \(\theta = \cos^{-1} x\) |
| Pythagorean Identity | Relates sine and cosine functions squared. | \(\sin^2 \theta + \cos^2 \theta = 1\) |
Additional Information: Relating Inverse Trig Functions
For angles \(\theta\) in the appropriate domains, we can relate inverse trigonometric functions using identities derived from right-angled triangles or fundamental identities.
If we have \(\cos \theta = x\), we can construct a right-angled triangle where the adjacent side is \(x\) and the hypotenuse is \(1\). The opposite side would be \(\sqrt{1-x^2}\).
In our case, \(\cos \theta = \frac{3}{\sqrt{17}}\).
We can consider a right-angled triangle with adjacent side \(3\) and hypotenuse \(\sqrt{17}\). The opposite side \(o\) can be found using the Pythagorean theorem: \(3^2 + o^2 = (\sqrt{17})^2\), which gives \(9 + o^2 = 17\), so \(o^2 = 17 - 9 = 8\), and \(o = \sqrt{8} = 2\sqrt{2}\).
For the same angle \(\theta\), \(\sin \theta\) would be \(\frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{8}}{\sqrt{17}}\).
Therefore, \(\sin^2 \theta = \left(\frac{\sqrt{8}}{\sqrt{17}}\right)^2 = \frac{8}{17}\).
This confirms our earlier calculation for \(\sin^2 \theta\) using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\).
The expression \(1+\sin ^2\left(\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\right)\) evaluates to \(1 + \frac{8}{17} = \frac{25}{17}\).
Important Questions from Inverse Trigonometric Functions
What is 2 cot \(\left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)\) equal to ?
If sec-1 p - cosec-1q = 0, where p > 0, q > 0; then what is the value of p-2 + q-2 ?
Consider the following statements:
1. There exists \({\rm{\theta }} \in \left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\) for which tan -1 (tan θ) ≠ θ
2. \({\sin ^{ - 1}}\left( {\frac{1}{3}} \right) - {\sin ^{ - 1}}\left( {\frac{1}{5}} \right) = {\sin ^{ - 1}}\left( {\frac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{15}}} \right)\)
Which of the above statements is/are correct?
Consider the following statements:
1. \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\rm{\pi }}\)
2. There exist x, y ∈ [-1, 1], where x ≠ y such that sin -1 x + cos -1 \({\rm{y}} = \frac{{\rm{\pi }}}{2}\)
Which of the above statements is/are correct?The value of \({\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)\) is
