document.write( "Question 709118: find the exact value of each expression\r
\n" ); document.write( "\n" ); document.write( "20) inverse cos( inverse sin root 2 over 3) \n" ); document.write( "
Algebra.Com's Answer #436444 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
If you posted the actual question you have then it is a bad question for several reasons:
  • What is \"root\"? Square root? cube root? 4th root? etc.
  • What is inside the radical (or whatever kind of root it is)? Just the 2? or 2/3?
  • The inverse functions take a ratio as input and provide an angle as output. Depending on what \"root 2 over 3\" means, it may be a valid ratio to be input to the inverse sin. But inverse sin will output an angle which is not valid input to inverse cos.
If what you posted is not literally the problem given to you (which is waht I suspect), then please:
  • Be more careful in posting. There is virtually no chance an inverse sin is input to an inverse cos.
  • Don't just say \"root\". Tell us what kind of root.
  • Use parentheses to tell us what is inside the radical of the root. Use
    \n" ); document.write( "square root(2)/3 (or just sqrt(2)/3) for \"sqrt%282%29%2F3\"
    \n" ); document.write( "and use
    \n" ); document.write( "square root(2/3) (or just sqrt(2/3)) for \"sqrt%282%2F3%29\"
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