Expert: Steve Holleran - 7/25/2008

Question
Can you please help me to solve these three math problems?

a) Give the value of  tan[tan^-1 ((x+1)/(x-1)) + tan^-1 ((x-1)/(x))]

b) Solve cos^-1(x) + cos^-1(2x) = 60 degrees

c) Verify the identity: tan(2tan^-1(x)) = 2tan[tan^-1(x)+tan^-1(x^3)]


thank  you

Answer
Hi Andy,

A)  Here, call tan^-1((x+1)/(x-1)) = A   and
              tan^-1((x-1)/(x))   = B

Then you have tan[A + B].  Use the identity

tan[A + B] = (tan A + tan B)/(1 - tan A tan B) and get

          =  [(x+1)/(x-1) + (x-1)/x]  / [1 - (x+1)/(x-1) * (x-1)/x]

This simplifies to [(2x^2 -x + 1)/x(x-1)] / (-1/x)

or [2x^2 -x +1]/(1-x)


B)  Do likewise here:  A = cos^-1 (x) and B = cos^-1 (2x).

   So the equation becomes A + B = 60 deg

now take the cosine of each side:

  cos(A + B) = cos 60 = 1/2

  cos A cos B - sin A sin B = 1/2

 (x)(2x) - (rt(1-x^2))(rt(1-4x^2)) = 1/2

   2x^2 - rt[1 - 5x^2 + 4x^4] = 1/2

   4x^2 - 2rt[1-5x^2+4x^4] = 1

   4x^2 -1 = 2rt[1-5x^2+4x^4]

Square each side:

   16x^4 -8x^2 + 1 = 4(1-5x^2+4x^4)

   16x^4 -8x^2 + 1 = 4-20x^2+16x^4

         12x^2 - 3 = 0

          4x^2 - 1 = 0

    (2x+1)(2x-1)=0    so x = +/- 1/2


C)  Call tan^-1 x = A and tan^-1 x^3 = B

  Left side:  tan[2tan^-1 x] = tan [2A]

                             = (2 tan A) / (1-tan^2 A)

                             = (2x) / (1-x^2)


  Right side:  2 tan[tan^-1 x + tan^-1 x^3] = 2[tan A + tan B]

              = 2* [ (tan A + tan B) / (1 - tan A tan B)]

              = 2 * [(x + x^3) / (1 - x * x^3)]

              = [2x(1+x^2) / (1-x^4)]

              = [2x(1 + x^2) / (1 + x^2)(1 - x^2)]

              = (2x) / (1 - x^2)


I hope this helps out.
Steve