Question 820355
{{{cot^-1(u)}}} is a reference to an angle. Specifically it is an angle whose cot ratio is "u" or {{{u/1}}}<br>
It will probably help if you have a drawing:<ol><li>Draw a right triangle.</li><li>Choose one of the acute angles and label it A. This will be the angle represented by {{{cot^-1(u)}}}.</li><li>We want cot(A) to be {{{u/1}}}. 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.</li></ol>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:
{{{u^2 + 1^2 = h^2}}} (where "h" stands for the hypotenuse)
Simplifying:
{{{u^2 + 1 = h^2}}}
Square root of each side. (And since the hypotenuse is never negative, we will use only the positive square root (and not use a <u>+</u> as usual).
{{{sqrt(u^2 + 1) = sqrt(h^2)}}}
{{{sqrt(u^2 + 1) = h}}}<br>
Now we can find the sin (opposite/hypotenuse):
{{{sin(cot^-1(u)) = 1/sqrt(u^2+1)}}}
This may be an acceptable answer.<br>
But it does have a square root in the denominator. So we may want to rationalize it:
{{{sin(cot^-1(u)) = (1/sqrt(u^2+1))(sqrt(u^2+1)/sqrt(u^2+1))}}}
{{{sin(cot^-1(u)) = sqrt(u^2+1)/(u^2+1)}}}