Cos(arcsin(-1/3)) how to do it by hand?

[HappMatt]

cos(arcsin(-1/3)) how to do it by hand??

Homework Statement

Solve without a calculator
cos(arcsin(-1/3))

The Attempt at a Solution

now i have fount the solution to be (-2/3)(2^(1/2)) now not only did my calculator come up with that but between the SSS formula and pythagorean formula( i think is what we cal it, a^2+b^2...) i was able to get the same answer but the question says not to use a calculator and using those formulas of course I am going to have to use a calculator to figure out some of the roots and what not.

I haven't had to do this stuff for quite sometime now I am a ways past precalc but i was trying to help a friend and I am not sure how to do this without a calculator or a root table and any other aids other than my mind. So being that this is precalc i would assume there is a easy way do do this beyond what I am trying if the teacher expects it to be done without a calc so if you know how please enlighten me.

 

[Feldoh]

It's actually very simple once you know how to do it. Forget about cosine for a moment. Some angle (x) equals the arcsine of -1/3 -- so x=arcsin(-1/3). So you know that the sin(x)=-1/3. Now you can make a right triangle in the IV quadrant and solve for the cosine value, since you are simply solving for the cos(arcsin(-1/3)) or cos(x).

 

[VietDao29]

You can also try to work out the general formula for cos(arcsin(x)). Here we go.
Let \alpha = \arcsin(x) \Rightarrow \sin \alpha = x
cos(arcsin(x)) will become: \cos \alpha

Also, you should note that arcsin(x) will return the values in the interval \left[ - \frac{\pi}{2} ; \ \frac{\pi}{2}\right], i.e \alpha is in I, and IV quadrant, and its cosine value should be positive. So we have:
\cos \alpha = \sqrt{1 - \sin ^ 2 \alpha} = \sqrt{1 - x ^ 2}

Applying the formula to your question yielding:
\cos \left( \arcsin \left( - \frac{1}{3} \right) \right)) = \sqrt{1 - \left( -\frac{1}{3} \right) ^ 2} = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}. :)

You can do the same and find the general formula to:
sin(arccos(x))
Can you get it? :)

 

[HappMatt]

You can also try to work out the general formula for cos(arcsin(x)). Here we go.
Let \alpha = \arcsin(x) \Rightarrow \sin \alpha = x
cos(arcsin(x)) will become: \cos \alpha

Also, you should note that arcsin(x) will return the values in the interval \left[ - \frac{\pi}{2} ; \ \frac{\pi}{2}\right], i.e \alpha is in I, and IV quadrant, and its cosine value should be positive. So we have:
\cos \alpha = \sqrt{1 - \sin ^ 2 \alpha} = \sqrt{1 - x ^ 2}

Applying the formula to your question yielding:
\cos \left( \arcsin \left( - \frac{1}{3} \right) \right)) = \sqrt{1 - \left( -\frac{1}{3} \right) ^ 2} = \sqrt{1 - \frac{1}{9}} = \frac{2\sqrt{2}}{3}. :)

You can do the same and find the general formula to:
sin(arccos(x))
Can you get it? :)

thanks for both of you for the replies. I especially apreciate yours VietDao29, that way seems to be the best route in finding the aswer via no calculator because of its simplicity, i know with the triangle method although easy not so much without the calculator.

 

[Feldoh]

thanks for both of you for the replies. I especially apreciate yours VietDao29, that way seems to be the best route in finding the aswer via no calculator because of its simplicity, i know with the triangle method although easy not so much without the calculator.

Yeah I think his method is better too.