Chapter 1: Problem 37
Draw the graph of \(y=\pi / 2-\arcsin x\). Make a conjecture. Prove it.
Short Answer
Expert verified
The graph y=\(\pi/2-\arcsin x\) reflects \(\arcsin x\) across \(y=\pi/2\).
Step by step solution
01
Identify the Range and Domain
The function \(y = \pi/2 - \arcsin x\) involves the inverse sine function, \(\arcsin x\), which is defined only for \(x\) in the interval \([-1, 1]\). Therefore, the domain of the function is \([-1, 1]\). The range of \(\arcsin x\) is \([-\pi/2, \pi/2]\), so the range of \(y = \pi/2 - \arcsin x\) will be \([0, \pi]\).
02
Analyze the Transformation of the Graph
The function \(y = \arcsin x\) is flipped vertically and shifted by \(\pi/2\) because of the expression \(\pi/2 - \arcsin x\). Therefore, it will invert the values of the \(\arcsin x\) graph and shift them upward by \(\pi/2\) units.
03
Sketch the Graph
To draw the graph, start by considering key points from \(y = \arcsin x\). For \(x = -1\), \(\arcsin(-1) = -\pi/2\), making \(y = \pi/2 - (-\pi/2) = \pi\). For \(x = 0\), \(\arcsin(0) = 0\), making \(y = \pi/2 - 0 = \pi/2\). For \(x = 1\), \(\arcsin(1) = \pi/2\), making \(y = \pi/2 - \pi/2 = 0\). Plot these points and draw a smooth curve between them to complete the graph. The graph is decreasing on \([-1, 1]\).
04
Make a Conjecture
Based on the graph, one can conjecture that \(y = \pi/2 - \arcsin x\) is the reflection of \(\arcsin x\) across the line \(y = \pi/2\).
05
Prove the Conjecture
To prove the conjecture, consider that if \(y = \pi/2 - \arcsin x\), then rearranging gives \(\arcsin x = \pi/2 - y\). Plugging this into the equation yields \(x = \sin(\pi/2 - y)\). Using the trigonometric identity, \(\sin(\pi/2 - y) = \cos(y)\), we can express \(x = \cos(y)\). Therefore, the function indeed transforms points as hypothesized, confirming the reflection about the line \(y = \pi/2\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graph Analysis
The graph of the function \(y = \frac{\pi}{2} - \arcsin x\) is intriguing, as it involves an inverse trigonometric function. When analyzing this graph, we start by identifying crucial points derived from the function \(y = \arcsin x\). \(y = \arcsin x\) itself is defined for \(x\) in the range -1 to 1, moving from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).
\(y = \frac{\pi}{2} - \arcsin x\) repositions these values, reflecting them vertically about the line \(y=\frac{\pi}{2}\). This means the graph starts at \(\pi\) when \(x=-1\), descends to \(\frac{\pi}{2}\) at \(x=0\), and reaches 0 at \(x=1\). The graph's smooth curve signifies a consistent rate of change, gradually decreasing across its domain.
\(y = \frac{\pi}{2} - \arcsin x\) repositions these values, reflecting them vertically about the line \(y=\frac{\pi}{2}\). This means the graph starts at \(\pi\) when \(x=-1\), descends to \(\frac{\pi}{2}\) at \(x=0\), and reaches 0 at \(x=1\). The graph's smooth curve signifies a consistent rate of change, gradually decreasing across its domain.
Function Transformation
The function \(y = \frac{\pi}{2} - \arcsin x\) involves both reflection and vertical translation. The expression \(\frac{\pi}{2} - \arcsin x\) can be broken down into two operations:
- A vertical shift: The constant \(\frac{\pi}{2}\) shifts the graph of \(\arcsin x\) upwards by \(\frac{\pi}{2}\) units.
- A vertical reflection: By subtracting \(\arcsin x\) from \(\frac{\pi}{2}\), the graph experiences a reflection across the line \(y = \frac{\pi}{2}\).
Domain and Range
Understanding the domain and range of a function is crucial in graphing inverse trigonometric functions like \(y = \frac{\pi}{2} - \arcsin x\). In this case, the domain is determined by the arcsine function, \(-1 \leq x \leq 1\).
The range for \(\arcsin x\) being \([-\frac{\pi}{2}, \frac{\pi}{2}]\), results in the transformed range for \(y = \frac{\pi}{2} - \arcsin x\) being \([0, \pi]\).
The range for \(\arcsin x\) being \([-\frac{\pi}{2}, \frac{\pi}{2}]\), results in the transformed range for \(y = \frac{\pi}{2} - \arcsin x\) being \([0, \pi]\).
- The lowest output of the function when \(x = 1\) is 0.
- Its highest output when \(x = -1\) is \(\pi\).
Trigonometric Identities
Trigonometric identities assist dramatically in interpreting transformations within inverse trigonometric functions. In \(y = \frac{\pi}{2} - \arcsin x\), our focus shifts to the identity \(\sin(\frac{\pi}{2} - y) = \cos(y)\). This identity confirms our conjecture about the reflection by rewriting the equation for transformation.
Starting with \(y = \frac{\pi}{2} - \arcsin x\), we rearrange to \(\arcsin x = \frac{\pi}{2} - y\). Substituting this back to the definition of sine, we find \(x = \cos(y)\), illustrating that the sine and cosine identity achieves our desired mirroring effect.
Starting with \(y = \frac{\pi}{2} - \arcsin x\), we rearrange to \(\arcsin x = \frac{\pi}{2} - y\). Substituting this back to the definition of sine, we find \(x = \cos(y)\), illustrating that the sine and cosine identity achieves our desired mirroring effect.
- This confirms \(x\)'s output equivalent to cosine's transformation from sine's inverse.
- It offers profound insight into how trigonometric functions complement each other through symmetric properties.