I don't get this one still...?

Question

Thx for the answers on the others, but plz explain these in more detail. I need to get rid of the subtraction...only addition, multip and division are allowed

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

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

Answer 1

To do sin(a+x) - sin(a)

you need to know the expansion

sin(a+b) = sin(a)cos(b) + cos(a) sin(b)

and similar expansions

Using this with similar expansions it's possible to show that

sin S - sin D = 2 sin((S-D)/2) cos((S+D)/2)

That gives

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

Then when you put it over x, and use the fact that

limit of [sin(x/2) / (x/2)] as x --> 0 is 1, you find the limit of the expression is cos a.

d)  This one relates to the factorisation of p^3 - q^3 into

(p - q)(p^2 + pq - q^2)

If p = (1+x)^(1/3) and q = 1, then p^3 = 1+x and the above factorisation becomes

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

Let's write D = ((1+x)^(2/3) + (1+x)^(1/3) + 1)

Then the above statement becomes

x = ((1+x)^(1/3) - 1)D

Therefore

((1+x)^(1/3) - 1) = x/D

The point of this is that when we divide this by x we get 1/D,

and then if we let x --> 0

D --> 1 + 1 + 1

i.e. 1/D --> 1/3

Answer 2

For the first one, I googled "trignometric identities" and found this useful one:

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

Applied to your case, you get:

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)

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To make "THE Nerd"'s answer below a little more clear:

If you have:  r³ - t³ = (r - t)(r² - rt - t²)

then you can rearrange this to get:  (r - t) = (r³ - t³) / (r² - rt - t²)

Answer 3

Since he did c, I'll do d

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