The exact value of \( y = \cos\left(\arcsin\frac{5}{11}\right) \) can be determined using the Pythagorean identity \( 1 = \cos^2(\theta) + \sin^2(\theta) \). Given that \( \sin(\theta) = \frac{5}{11} \), we find \( \cos(\theta) \) by rearranging the identity to \( \cos^2(\theta) = 1 - \left(\frac{5}{11}\right)^2 \). This leads to \( \cos(\theta) = \sqrt{1 - \frac{25}{121}} = \sqrt{\frac{96}{121}} = \frac{4\sqrt{6}}{11} \). The discussion emphasizes the importance of recognizing that \( \theta = \arcsin\frac{5}{11} \) is restricted to the first quadrant, where cosine is positive.
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Just to test your understanding please tell me the unrestricted domain of the sine function. When the domain is not restricted is it one-one or many one function? Also state the definition of a function using all terms such as "domain" correctly.opus said:Ok thank you for responding. That is more clear and precise. I felt myself rambling and talking in circles trying to explain it, but your edits are much better. Thank you again!
The graph of the sine function goes on forever in both the positive and negative directions and we can make as many positive or negative unit circle rotations as we’d like. For these reasons the unrestricted domain of the sine function is[-∞,∞]. This would be a one to many function where an infinite amount of inputs would map to a single output. For example, (π/2) and (-3π/2) would both map to 1.Let'sthink said:Just to test your understanding please tell me the unrestricted domain of the sine function. When the domain is not restricted is it one-one or many one function? Also state the definition of a function using all terms such as "domain" correctly.
Again you are almost right and like earlier have made some terminological mistakes! I correct them by writing in red. Your definition is almost correct. But just I write it again in so called my words. Function is a rule which associates with EVERY element in the domain set, with a UNIQUE element in the Range set. That is why a function can be one one or many one (but never one many, because then it will not be function at all) So the inverse of a many one function cannot be a function unless we restrict the range set and make it one one. This problem illustrates these ideas.opus said:The graph of the sine function goes on forever in both the positive and negative directions and we can make as many positive or negative unit circle rotations as we’d like. For these reasons the unrestricted domain of the sine function is[-∞,∞]. This would be a many to one function where an infinite number of inputs would map to a single output in a limited range. For example, (π/2) and (-3π/2) would both map to 1.
For the last part, do you mean something like f is a function that maps an input,x, to one and only one output, y, for all values of x in the domain of f? I didnt look this one up because I wanted to attempt to define it on my own to test my understanding. It doesn't seem wholly sufficient.
Ah that is much more precise. I definitely feel much better about my understanding of functions after going over this. Thank you so much.Let'sthink said:Again you are almost right and like earlier have made some terminological mistakes! I correct them by writing in red. Your definition is almost correct. But just I write it again in so called my words. Function is a rule which associates with EVERY element in the domain set, with a UNIQUE element in the Range set. That is why a function can be one one or many one (but never one many, because then it will not be function at all) So the inverse of a many one function cannot be a function unless we restrict the range set and make it one one. This problem illustrates these ideas.