Are There Multiple Solutions for Sin(x) and Arcsin(x) at Pi/4?

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For the problem sin-(1/sqrt2) ...(sin-.. being arcsin) the answer is pi/4 but is that the only answer becuase pi/4 lies between [-90,90]?


or would it also be right to say PI-4PI=3pi/4? (although i think this is wrong)
 
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Do you know what the graph of sin(x) looks like? It goes up from (0,0) to (\pi/2, 1) then back down to (\pi,0). You are right that sin(3\pi /4) is also equal to \sqrt{2}{2} but IF YOUR PROBLEM SPECIFIES THAT THE ANSWER MUST BE BETWEEN -\pi /2 to \pi/2. If it does not then 3\pi /4 is the &quot;principal&quot; value (it&#039;s the value your calculator gives you) so if you are asking that &quot;tan<sup>-1</sup>&quot; be a single valued function, that would be its value. If you are solving &quot;tan(x)= \sqrt{2}/2&quot; then there are an infinite number of solutions- rhe two you give plus any multiple of 2 \pi.
 
No the problem does not give the intervals of [-90,90]...just the question... so i can come to the conclusion that 3pi/4 is also correct (as is pi/4)
 
Halls.. My calculator gives me the answer of pi/4 though not 3pi/4 and where does the sqrt of 22 come from?
 
My "tex" messed up. It should have been \sqrt{2}/2.
Also the 3\pi /4 was just a typo on my part. I meant, of course, \pi/4.
By the way- it is really bad practice to talk about "intervals [-90,90]" AND give values in terms of \pi. You are going to have to choose whether you are working in degrees or radians! (I strongly recommend radians.)

Again, if your problem is to find all solutions to sin(x)= \sqrt{2}/2, then the solutions are all numbers of the form \pi/4 + 2n\pi and 3\pi/4+ 2n\pi where n is any integer. If your problem is to find Sin^{-1}(\sqrt{2}/2) with arcsine as a single-valued function, then the only answer is \pi/4. (Notice the capital "S" on "Sin-1". Many texts use the capital when they want to mean the single-valued function: the principal value.)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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