Problem 9
Question
Let \(m=\sin ^{-1}\left(a^{6}+1\right)+\cos ^{-1}\left(a^{4}+1\right)-\tan ^{-1}\left(a^{2}+1\right)\), then find the image of the line \(x+y=m\) about the \(y\)-axis.
Step-by-Step Solution
Verified Answer
The image of the line \(x + y = m\) about the y-axis is \(y = x + m\). This is achieved by changing the sign of all 'x' coordinates in the original equation to get the equation for the mirrored line.
1Step 1: Trigonometric Expression
Firstly, one can observe that there is a given expression \( m = sin^{-1}(a^{6}+1) + cos^{-1}(a^{4}+1) - tan^{-1}(a^{2}+1)\). This expression merely provides a common value for all the inverse trigonometric expressions. The inverse trigonometric functions are real-number functions, so they have to be cast into a form applicable for the problem. Thus, a new expression for 'm' gives us a line in terms of x and y.
2Step 2: Equation of the Line
The given equation of the line is \(x + y = m\). In order to plot this line and consequently mirror it about the y-axis, it's best to write it in slope-intercept form (y = mx + b). Subtracting 'x' from both sides of the equation gives \(y = -x + m\)
3Step 3: Reflect the Line about the y-axis
The line \(y = -x + m\) has all its 'x' coordinates changed to their opposites when mirrored about the y-axis. That means, the image of the line about the y-axis has the equation \(y = x + m\)
Key Concepts
Reflection of LinesCoordinate GeometryTrigonometric Identities
Reflection of Lines
Understanding the reflection of lines can be a helpful visualization exercise in coordinate geometry. When we talk about reflecting a line, we usually mean flipping it over one of the coordinate axes. Here, we reflect the line about the y-axis.
Reflecting a line across the y-axis involves changing the x-coordinates of every point on the line to their opposites. To illustrate, if we have a line described by the equation \(y = -x + m\), its reflection across the y-axis results in a new line equation \(y = x + m\).
This means:
Reflecting a line across the y-axis involves changing the x-coordinates of every point on the line to their opposites. To illustrate, if we have a line described by the equation \(y = -x + m\), its reflection across the y-axis results in a new line equation \(y = x + m\).
This means:
- Positive x-values become negative.
- Negative x-values become positive.
Coordinate Geometry
Coordinate geometry, often called analytic geometry, allows us to represent geometric shapes using coordinates and algebraic equations. It's like a bridge connecting algebra and geometry, where we can express geometric problems in a visual way.
Key components of coordinate geometry include:
Key components of coordinate geometry include:
- Points: Given as coordinates \((x, y)\).
- Lines: Typically represented as equations like \(y = mx + b\), where \(m\) is the slope.
- Equation of a Line: The equation \(x + y = m\) transforms to \(y = -x + m\) for simplicity.
Trigonometric Identities
Trigonometric identities involve relationships among trigonometric functions that hold true for all values within their domains. In the context of inverse trigonometric functions, these identities help simplify complex expressions.
Some important inverse trigonometric functions are:
The exercise involves these functions:
Some important inverse trigonometric functions are:
- \(\sin^{-1}(x)\)
- \(\cos^{-1}(x)\)
- \(\tan^{-1}(x)\)
The exercise involves these functions:
- \(m = \sin^{-1}(a^6 + 1) + \cos^{-1}(a^4 + 1) - \tan^{-1}(a^2 + 1)\)
Other exercises in this chapter
Problem 8
Find the value of \(\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)-\tan ^{-1}\left(\frac{\sqrt{5-2 \sqrt{6}}}{1+\sqrt{6}}\right)\)
View solution Problem 8
The smallest and the largest values of \(\tan ^{-1}\left(\frac{1-x}{1+x}\right), 0 \leq x \leq 1\) are (a) \(0, \pi\) (b) \(0, \frac{\pi}{4}\) (c) \(-\frac{\pi}
View solution Problem 9
The equation \(\sin ^{-1} x-\cos ^{-1} x=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) has (a) No solution (b) Unique Solution (c) Infinite No of soln (d) None
View solution Problem 10
If \(-\pi \leq x \leq 2 \pi\), then \(\cos ^{-1}(\cos x)\) is (a) \(x\) (b) \(\pi-x\) (c) \(2 \pi+x\) (d) \(2 \pi-x\)
View solution