Write an algebraic expression for the following:
sin(arctan〖(x+1)/(x-1)〗 )
draw your triangle. If the legs of the triangle are x+1 and x-1, the hypotenuse is √((x+1)^2 + (x-1)^2)) = √(2x^2+2)
so, sin(arctan((x+1)/(x-1)) = (x+1)/√(2x^2+2)
To write an algebraic expression for sin(arctan((x+1)/(x-1))), we need to use some trigonometric identities.
Let's start by using the identity tan(arctan(u)) = u. In our expression, arctan((x+1)/(x-1)) is the angle, so we can replace it with u. Now we have:
sin(arctan((x+1)/(x-1))) = sin(u)
Next, let's use the Pythagorean identity sin^2(u) + cos^2(u) = 1. Divide both sides of this equation by cos^2(u) to get:
sin^2(u)/cos^2(u) + cos^2(u)/cos^2(u) = 1/cos^2(u)
Using the trigonometric identity tan^2(u) + 1 = sec^2(u), we can change the expression on the left side to:
tan^2(u) + 1 = sec^2(u)
(sin^2(u) / cos^2(u)) + 1 = sec^2(u)
Since tan(u) = (x+1)/(x-1), we can substitute this value in:
[((x+1)/(x-1))^2 / cos^2(u)] + 1 = sec^2(u)
Now, recall that sec^2(u) = 1/(cos^2(u)), so we can rewrite our expression as:
[((x+1)/(x-1))^2 / cos^2(u)] + 1 = 1/(cos^2(u))
Multiplying both sides by cos^2(u) gives us:
[((x+1)/(x-1))^2] + cos^2(u) = 1
Finally, we can replace cos^2(u) with 1 - sin^2(u) using the Pythagorean identity:
[((x+1)/(x-1))^2] + 1 - sin^2(u) = 1
Since sin(u) is the expression we want, we can write our final algebraic expression as:
sin(arctan((x+1)/(x-1))) = √[1 - ((x+1)/(x-1))^2]