Inverse Trig Problem: Prove arcsin(1/x) = -arccos((sqrt{x^2 - 1})/x)

[apple2357]

Can anyone offer any thoughts on this problem please?

 

[apple2357]

What i have tried so far:

siny =1/x and cos y = - sqrt(1-1/x^2)

Then i am thinking there are conditions on siny and cosy which would lead to inequalities:

-1<=siny<=1 which leads to -1<=1/x<=1

and similarly for the cos y expression

And this is where i get stuck

I am also thinking of using the Pythag. identity ( sin squared + cos squared =1) but not sure where that leads to anything useful.
These are my thoughts so far but i would appreciate a hint of some sort please?
Thanks

 

[khansaheb]

Can anyone offer any thoughts on this problem please?

Please offer your thoughts - so that we know where to begin?

Did you try substitution?

 

[khansaheb]

If I were to solve this problem, I would substitute

x = sec (Θ)

 

[Dr.Peterson]

What i have tried so far:

siny =1/x and cos y = - sqrt(1-1/x^2)

Well, we don't know that both sides are equal; we want to prove they are. So you can't take this as given.

But you're right that we want to show that, if sin(y) = 1/x, with x <= -1, then cos(-y) = -sqrt(x^2-1)/x, and the red negative can be dropped.

I would immediately think about quadrants as well: Where must our angle y be?

Then I'd draw a triangle with angle y, and fill in sides. (If you prefer, you can use this just to guide your thinking about Pythagoras.)

Finally, I'd make sure the RHS is in the right quadrant.

If I were to solve this problem, I would substitute

x = sec (Θ)

That's basically what he's doing, but calling the angle y.

 

[apple2357]

Thanks for your thoughts so far. I will have a play.

 

[apple2357]

Ok, i have had a go: If anyone can check it over i would be grateful. The tricky bit is knowing which cos y to pick and i think i can justify it, but would prob fail under intense interrogation. I think we would pick the negative sqrt because we need to cos y to be positive?

 

[Dr.Peterson]

I think we would pick the negative sqrt because we need to cos y to be positive?

You found that y is in quadrant IV. What quadrant will -y be in? Therefore, what is the sign of cos(-y)?

On the other hand, given the sign of x, is it really true that [imath]\sqrt{x^2}=x[/imath]?

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