Solving Inverse Trig Function: sin-1(1/sqrt2) Without a Calculator

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MacLaddy
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Homework Statement



Find the exact value of this expression in degrees without using a calculator or table.

sin-1(1/sqrt2)



Homework Equations



Typical inverse function



The Attempt at a Solution



I can figure this out easily on my calculator, as it comes to 45 degrees, but how could I figure this out without one? It isn't on the unit circle.
 
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MacLaddy said:

Homework Statement



Find the exact value of this expression in degrees without using a calculator or table.

sin-1(1/sqrt2)



Homework Equations



Typical inverse function



The Attempt at a Solution



I can figure this out easily on my calculator, as it comes to 45 degrees, but how could I figure this out without one? It isn't on the unit circle.

Let y = sin-1(1/sqrt(2)). Then sin(y) = sin(sin-1(1/sqrt(2))) = 1/sqrt(2).

Can you figure it out from there?
 
No, I think I can follow you that far, and I can see that corresponds with a triangle and the Pythagorean theorem if I draw it out, but I am still not seeing how to end at 45 degrees.

EDIT* Oh, I may see. So I multiply everything on the triangle by the square root of 2, and that gives me sqrt2/2?
 
So far in my class we have been dealing mainly with either the unit circle, or a calculator. A few random identities in between, but nothing quite like this.

I have one more follow up question to that one, if I may, and it's because I think it's relevant to how the one above was completed. (if I did in fact figure that out correctly) The question below goes like this.

Find the exact value of each composition without using a calculator or table.

sec(arcsin(1/sqrt2)).

Trying this I can not find an exact value, other than sec(45deg), or 1/cos(45deg)
 
MacLaddy said:
Trying this I can not find an exact value, other than sec(45deg), or 1/cos(45deg)

You're right there. what is cos(45)? And what is the value of that 1/cos(45)
 
Blu3eyes said:
You're right there. what is cos(45)? And what is the value of that 1/cos(45)

Ugh, 2sqrt2/2. I must need sleep. Thanks Blu3eyes and Mark44. I appreciate it. Couldn't see the forest through the trees on that one.