Can someone help me with this algebra?

Question

In all of the exaamples below, I need to get rid of subtraction. In some you have to mult by the conjugate, etc...but the final answer cant have subtraction

a) (1-cosx) / x^2

b) log(x+1) - log(x) , x large

c)sin(a+x) - sin(a)

d) (1+x)^1/3 - 1

Answer 1

The first two answers are correct for a and b, but not c:

a) (1-cosx) / x²  ==> Multiply by 1 in the form of (1+cosx) / (1+cosx)

1 - cos²x / x²(1+cosx)

Remembering the identity, sin²x + cos²x = 1

1-cos²x = sin²x

sin²x / x²(1+cosx)     <~~ Answer

b) When logs are subtracted, you can take the expression into the log and divide it.

log(m) - log(n) = log(m/n)

log(x+1) - log(x)

= log [ (x+1) / x ]  <~~ Answer

c) sin u - sin v = 2 cos ((u + v)/2) sin ((u - v)/2)

sin (a+x) - sin (a) = 2 cos ((a+x + a)/2) sin ((a+x - a)/2)

sin (a+x) - sin (a) = 2 cos ((2a+x)/2) sin (x/2) <~~Answer

d) r³ - t³ = (r - t)(r² - rt - t²)

Like the first guy said, let r = (1+x)^(1/3) , t = 1

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

= x / (1+x)^2/3 + (1+x)^1/3 + 1  <~~Answer

Answer 2

a) (1-cosx)/ x^2 = 2 sin^2 x / x^2

b) log{(x+1)/x}

c) 2 sin (a+x/2) cos (x/2)

d)   1+1/3x-2/9x^2 +.........-1

   = 1/3x-2/9x^2+...............

Answer 3

(a) multiply by the conjugate.



    (1 - cos^2 x)/[x^2(1 + cos x)]

   = sin^2 x /[x^2(1 + cos x)]

(b) use the quotient rule for logs:

     log [(x+1)/x]

(c) sin (a + x) - sin (a) = 2 cos (2a + x) sin (x)

(d) if u remember the formula:

     a^3 - b^3 = (a-b)(a^2 + ab + b^2)  

  and u let a = (1 + x)^(1/3) and b = 1

u get (1+ x) - 1 on the left side which simplifies to x.