Tip:
Highlight text to annotate it
X
We rewrite sine of
arccotangent
of (2x)
as an algebraic expression.
We let t = arccotangent
of (2x).
Then cotangent of t equals (2x).
Then t
must be in the range of arccotangent, from 0 to pi, excluding the end points.
We are basically finding
sine of t.
And we substitute t
into the original.
We can rewrite 2x as 2x/1.
We let t prime, t', be the reference angle.
We draw a right triangle.
Cotangent, in this case, we can treat it as
adjacent over opposite.
So, we have 2x as the base of this right angle, one is the height.
We let h be the hypotenuse.
Using Pythagoren Theorem, h squared equals one squared plus
(2x) squared,
equals one
plus (4x) squared.
Take the square root on both sides, we have h = plus or minus the square root of (1 plus
(4x) squared).
Then, sine of t, we can treat it as sine of (t prime),
which is opposite over hypotenuse, one over
(plus or minus square root of (1 + 4 (x squared))).
We noticed that we keep the ± sign so that we are very careful to get the right answer.
We determined earlier that t is
in the interval from zero to pi,
which is in the first and second quadrants excluding the end-points.
Thus, sine of t must be positive.
The result is one over
(square root of (1 + 4 (x squared))).