trig reduction?

Question

reduce to an algebraic equation
cos(arcosx - arctanx)

Answer 1

cos ( acos x - atan x ) =
cos( acos x ) * cos (atan x) + sin( acos x ) * sin (atan x)



also u can simplify it by
cos(acos x) = x
sin (asin x) = x

where..
acos x = asin ( √(1- x² ) )
atan x =
atan x = (π/2) - acos ( x / √(1 + x²) )

so the equation would be
cos( acos x ) * cos (atan x) + sin( acos x ) * sin (atan x)

cos( acos x ) * sin (asin ( x / √(1 + x²) )) / tan(atan x) + sin( asin ( √(1- x² ) ) ) * sin (asin ( x / √(1 + x²) ))

as shown at the link (readable picture)
http://img218.imageshack.us/img218/1484/untitledmh6.jpg

Answer 2

Use the angle difference formula for cosine.

cos(arccosx - arctanx)
= [cos(arccosx)] [cos(arctanx)] + [sin(arccosx)] [sin(arctanx)]

= x [1 / √(1 + x²)] + √(1 - x²) [x / √(1 + x²)]

= x / √(1 + x²) + x√(1 - x²) / √(1 + x²)

= x[1 + √(1 - x²)] / √(1 + x²)

Answer 3

cos(A-B) = cos(A)cos(B) + sin(A)sin(B)
so...
cos(arcosx - arctanx) = cos(arccos(x))cos(arctan(x)) + sin(arccos(x))sin(arctan(x))

I like to solve these by drawing triangles. Imagine "theta" = arccos(x) by labeling the leg adjacent to theta as "x" and the hypoteneuse as "1". The other leg becomes sqrt(1 - x^2). So cos(arccos(x)) = x (obviously) and sin(arccos(x)) = sqrt(1 - x^2).

Similarly, cos(arctan(x)) = 1/(1 + x^2) while sin(arctan(x)) = x/(1 + x^2).

Sub into the formula and simplify if you like...