document.write( "Question 820355: Write the trigonometric expression sin (cot-1 u) as an algebraic expression in u. \n" ); document.write( "

Algebra.Com's Answer #493496 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"cot%5E-1%28u%29\"$ is a reference to an angle. Specifically it is an angle whose cot ratio is \"u\" or $\"u%2F1\"$

\n" ); document.write( "It will probably help if you have a drawing:

Draw a right triangle.
Choose one of the acute angles and label it A. This will be the angle represented by $\"cot%5E-1%28u%29\"$ .
We want cot(A) to be $\"u%2F1\"$ . Since cot is adjacent/opposite, label the side adjacent to A as \"u\" and the side opposite to A as \"1\". This makes cot(A) = u.

Now we want to find the sin(A). Since sin is opposite/hypotenuse and since we do not yet have the the hypotenuse, we must find the hypotenuse next. Using the Pythagorean Theorem we get:
\n" ); document.write( " $\"u%5E2+%2B+1%5E2+=+h%5E2\"$ (where \"h\" stands for the hypotenuse)
\n" ); document.write( "Simplifying:
\n" ); document.write( " $\"u%5E2+%2B+1+=+h%5E2\"$
\n" ); document.write( "Square root of each side. (And since the hypotenuse is never negative, we will use only the positive square root (and not use a + as usual).
\n" ); document.write( " $\"sqrt%28u%5E2+%2B+1%29+=+sqrt%28h%5E2%29\"$
\n" ); document.write( " $\"sqrt%28u%5E2+%2B+1%29+=+h\"$

\n" ); document.write( "Now we can find the sin (opposite/hypotenuse):
\n" ); document.write( " $\"sin%28cot%5E-1%28u%29%29+=+1%2Fsqrt%28u%5E2%2B1%29\"$
\n" ); document.write( "This may be an acceptable answer.

\n" ); document.write( "But it does have a square root in the denominator. So we may want to rationalize it:
\n" ); document.write( "

\n" ); document.write( " $\"sin%28cot%5E-1%28u%29%29+=+sqrt%28u%5E2%2B1%29%2F%28u%5E2%2B1%29\"$ \n" ); document.write( "

\n" );