Question

Write the trignometric expression as an algebraic expression: sin(arctan(2x)-arccos(x))

Answer 1

let arctan 2x = a
let arccos x = b

now we have sin (a - b) = sin a cos b - sin b cos a

since tan a = 2x = opp/adj, then opp = 2x and adj = 1
since opp^2 + adj^2 = hyp^2, we have (2x)^2 + 1 = 4x^2 + 1 = hyp^2 ==> hyp = sqrt(4x^2 + 1)
sin a = 2x / sqrt(4x^2 = 1) and cos a = 1 / sqrt(4x^2 + 1)

since cos b = x = adj / hyp, we have adj = x and hyp = 1
x^2 + opp^2 = 1 ==> 1 - x^2 = opp^2 ==> opp = sqrt(1 - x^2)
sin b = sqrt(1 - x^2) / 1 = sqrt(1 - x^2), and we already know that cos b = x

sub into the formula:
sin (a - b) = (2x / sqrt(4x^2 - 1))(x) - (1 / sqrt(4x^2+ 1))sqrt(1 - x^2)

and thus sin (arctan 2x - arccos x) = 2x^2 / sqrt(4x^2 - 1) - sqrt(1 - x^2) / sqrt(4x^2 + 1)
= (2x^2 - sqrt(1 - x^2)) / sqrt(4x^2 + 1)