evaluate the expression sin(tan^-1(x/2)+cos^-1(x)) and cos(2sin^-1(x))?

Answer 1

sin(tan^-1(x/2)+cos^-1(x))

let y = arctan(x/2) and z = arccos (x)
tan y = x/2 ==> sin y = x/√(x² + 4) and cos y = 2/√(x² + 4)
cos z = x ==> sin z = √(1 - x²) and tan z = √(1 - x²)/x

sin(tan^-1(x/2)+cos^-1(x))
= sin(y + z) = sin y cos z + cos y sin z
= x²/√(x² + 4) + 2√(1 - x²)/√(x² + 4)
= [x² + 2√(1 - x²)]/√(x² + 4) <<< answer

cos(2sin^-1(x))
Let y = arcsin x
sin y = x ==> cos y = √(1 - x²) and tan y = x/√(1 - x²)
cos(2sin^-1(x)) = cos (2y) = 2 (cos y)² - 1
= (2)(√(1 - x²))² - 1
= 2 (1 - x²) - 1
= 2 - 2x² - 1
= 1 - 2x² <<< answer

Answer 2

Let A = sin^(-1)(x), B = tan^(-1)(x/2) and C = cos^(-1)(x).

Then sin(A) = x, tan(B) = x/2, and cos(C) = x

cos(2A) = 1 - 2sin^2(A) = 1 - 2x^2.

sin(B + C) = sin(B) cos(C) + cos(B) sin(C)

draw a right triangle with angle B for one of the acute (less than 90) angles.

since tan(B) = x/2 = opp/adj, label the opp. side of angle B in the triangle x and the adj. side 2.

Using the Pythagorean theorem, the hypotenuse is sqrt(x^2 + 4)

so cos(B) = adj./hyp. = 2/sqrt(x^2 + 4), and sin(B) = opp/hyp = x/sqrt(x^2 + 4).

since cos(C) = x/1= adj/hyp, draw another right triangle for angle C with x = adj. side and 1 = hyp. Again using the Pythagorean thm., opp = \sqrt(1 - x^2).

Then sin(C) = opp/hyp = sqrt(1 - x^2)/1

Now just plug these into the equation above for sin(B + C):

sin(B + C) = sin(B) cos(C) + cos(B) sin(C)
= [ x/sqrt(x^2 + 4) ][ x ] + [ 2/sqrt(x^2 + 4) ][ sqrt(1 - x^2) ]

Answer 3

hmmmmmmmmmmmmmmmm
dosn't that depend on what x equals??