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

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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 im going to have to use a calculator to figure out some of the roots and what not.

I havent had to do this stuff for quite sometime now im a ways past precalc but i was trying to help a friend and im 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 im trying if the teacher expects it to be done without a calc so if you know how please enlighten me.
 

Answers and Replies

  • #2
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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).
 
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  • #3
VietDao29
Homework Helper
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You can also try to work out the general formula for cos(arcsin(x)). Here we go.
Let [tex]\alpha = \arcsin(x) \Rightarrow \sin \alpha = x[/tex]
cos(arcsin(x)) will become: [tex]\cos \alpha[/tex]

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

Applying the formula to your question yielding:
[tex]\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}[/tex]. :)

You can do the same and find the general formula to:
sin(arccos(x))
Can you get it? :)
 
  • #4
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You can also try to work out the general formula for cos(arcsin(x)). Here we go.
Let [tex]\alpha = \arcsin(x) \Rightarrow \sin \alpha = x[/tex]
cos(arcsin(x)) will become: [tex]\cos \alpha[/tex]

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

Applying the formula to your question yielding:
[tex]\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}[/tex]. :)

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.
 
  • #5
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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. :biggrin:
 
Last edited:

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