# Refresh my memory , Inverse Trig

Refresh my memory plz, Inverse Trig

$$x=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)$$

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

So $$x=-\frac{\pi}{4}$$

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dynamicsolo
Homework Helper
$$x=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)$$

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

So $$x=-\frac{\pi}{4}$$
Correct -- and your calculator and mine will back you up on that...

Correct -- and your calculator and mine will back you up on that...

$$\sin^{-1}\left(-\frac{1}{\sqrt 2}\right)\approx -0.785$$

$$\sin\left(-\frac{\pi}{4}\right)\approx -0.707$$

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.

dynamicsolo
Homework Helper

$$\sin^{-1}\left(-\frac{1}{\sqrt 2}\right)\approx -0.785$$

$$\sin\left(-\frac{\pi}{4}\right)\approx -0.707$$
And -45º converted to radians is? (In other words, what's $$\frac{-\pi}{4}$$?)

[I expect we're going to hear a "D'oh!" in 3, 2, 1, ...]

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

[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?

dynamicsolo
Homework Helper
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 :)