Refresh my memory , Inverse Trig

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Homework Help Overview

The discussion revolves around the properties and calculations involving inverse trigonometric functions, specifically the inverse sine function. Participants are examining the relationship between angles and their sine values, particularly focusing on the angle corresponding to \(\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)\).

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition range of the inverse sine function and its implications for determining angles. There are attempts to reconcile calculator outputs with theoretical values, leading to questions about the accuracy of calculations and the relationship between angles and their sine values.

Discussion Status

The conversation is active, with participants exploring different interpretations of the outputs from calculators and the mathematical relationships involved. Some guidance is offered regarding the expected outputs of the sine and inverse sine functions, but confusion remains about the calculator's behavior.

Contextual Notes

There is an underlying assumption that participants are familiar with radians and degrees, as well as the properties of trigonometric functions. The discussion hints at potential misunderstandings regarding the outputs of calculators when evaluating sine and inverse sine functions.

rocomath
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Refresh my memory please, Inverse Trig

[tex]x=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)[/tex]

Inverse sine is defined from [tex]-\frac{\pi}{2}[/tex] to [tex]\frac{\pi}{2}[/tex] which lies in the 1st and 4th quadrant.

So [tex]x=-\frac{\pi}{4}[/tex]
 
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rocomath said:
[tex]x=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)[/tex]

Inverse sine is defined from [tex]-\frac{\pi}{2}[/tex] to [tex]\frac{\pi}{2}[/tex] which lies in the 1st and 4th quadrant.

So [tex]x=-\frac{\pi}{4}[/tex]

Correct -- and your calculator and mine will back you up on that...
 
dynamicsolo said:
Correct -- and your calculator and mine will back you up on that...
That's the thing, it gives me different answers. (in radians)

[tex]\sin^{-1}\left(-\frac{1}{\sqrt 2}\right)\approx -0.785[/tex]

[tex]\sin\left(-\frac{\pi}{4}\right)\approx -0.707[/tex]
 
Those aren't different answers. -pi / 4 is -0.785 and -1/ root 2 is -0.707.

The inverse sin of the ratio will give you the angle, the sin of the angle will give you the ratio.
 
rocomath said:
That's the thing, it gives me different answers. (in radians)

[tex]\sin^{-1}\left(-\frac{1}{\sqrt 2}\right)\approx -0.785[/tex]

[tex]\sin\left(-\frac{\pi}{4}\right)\approx -0.707[/tex]

And -45º converted to radians is? (In other words, what's [tex]\frac{-\pi}{4}[/tex]?)

[I expect we're going to hear a "D'oh!" in 3, 2, 1, ...]
 
Last edited:
dynamicsolo said:
And -45º converted to radians is? (In other words, what's [tex]\frac{-\pi}{4}[/tex]?)

[I expect we're going to hear a "D'oh!" in 3, 2, 1, ...]
LOL, I know that ... but what's with the calculator? Am I putting it in right?
 
rocomath said:
LOL, I know that ... but what's with the calculator? Am I putting it in right?

Of course you are! The sine of -45º or -pi/4 radians (which is approximately -0.785398... radians) gives you -1/sqrt(2) =
-[sqrt(2)]/2 = -0.70710678... ,
so taking the arcsine of (-0.70710678...) should give you
-0.785398...

I guess we're puzzled why you're puzzled. There's no reason you should get the same number in both directions. The magnitude of arcsin(x) doesn't match the magnitude of sin(x) at x = (pi)/4 ...
 
LMAO ... omg, I crack myself up :)
 

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