I'm having a complete mind blank..and can't do these two questions...

any help would be greatly appreciated.

1) By writing sin3x as sin (2x+x) show that sin 3x = 3sinx - 4sin(cubed)x

i know that sin(2x+x) = sin2xcosx + cos2xsinx but i don't know where to go from there.

2) f(x) = 1- 3 + 3 x does not = -2
x+2 (x+2)squared

show that f(x) = x(squared) + x + 1 x does not = -2
(x+2)squared

thanks

2. Originally Posted by Oranges&Lemons
I'm having a complete mind blank..and can't do these two questions...

any help would be greatly appreciated.

1) By writing sin3x as sin (2x+x) show that sin 3x = 3sinx - 4sin(cubed)x

i know that sin(2x+x) = sin2xcosx + cos2xsinx but i don't know where to go from there.

2) f(x) = 1- 3 + 3 x does not = -2
x+2 (x+2)squared

show that f(x) = x(squared) + x + 1 x does not = -2
(x+2)squared

thanks
oops, number 2 didnt come out properly, the x+2 is supposed to be under the first 3 and the (x+2)squared is supposed to be under the second 3...and then the (x+2)squared is supposed to be under x(squared) + x + 1. sorry!

3. Originally Posted by Oranges&Lemons
1) By writing sin3x as sin (2x+x) show that sin 3x = 3sinx - 4sin(cubed)x

i know that sin(2x+x) = sin2xcosx + cos2xsinx but i don't know where to go from there.
You're almost there.
$sin(2x + x) = sin(2x)cos(x) + cos(2x)sin(x)$

And we know that
$sin(2x) = 2sin(x)cos(x)$
and
$cos(2x) = 1 - 2sin^2(x)$
and
$cos^2(x) = 1 - sin^2(x)$

You finish it from here.

-Dan

4. Originally Posted by Oranges&Lemons
2) f(x) = 1- 3 + 3 x does not = -2
x+2 (x+2)squared

show that f(x) = x(squared) + x + 1 x does not = -2
(x+2)squared
Yes, the forum word processor doesn't like extra spaces does it?

If I am reading this correctly it is:
$f(x) = 1 - \frac{3}{x + 2} + \frac{3}{(x + 2)^2}$; $x \neq -2$

and you need to show that
$f(x) = \frac{x^2 + x + 1}{(x + 2)^2}$; $x \neq -2$

I'm going to just drop the $x \neq -2$ part and just assume its there.

So we've got
$f(x) = 1 - \frac{3}{x + 2} + \frac{3}{(x + 2)^2}$

Let's add the fractions. The LCM of $1, x + 2, (x + 2)^2$ is, of course, $(x + 2)^2$, so
$f(x) = 1 \cdot \frac{(x + 2)^2}{(x + 2)^2} - \frac{3}{x + 2} \cdot \frac{x + 2}{x + 2}+ \frac{3}{(x + 2)^2}$

$f(x) = \frac{(x + 2)^2 - 3(x + 2) + 3}{(x + 2)^2}$

See if you can finish this from here.

-Dan

5. Originally Posted by topsquark
Yes, the forum word processor doesn't like extra spaces does it?

If I am reading this correctly it is:
$f(x) = 1 - \frac{3}{x + 2} + \frac{3}{(x + 2)^2}$; $x \neq -2$

and you need to show that
$f(x) = \frac{x^2 + x + 1}{(x + 2)^2}$; $x \neq -2$

I'm going to just drop the $x \neq -2$ part and just assume its there.

So we've got
$f(x) = 1 - \frac{3}{x + 2} + \frac{3}{(x + 2)^2}$

Let's add the fractions. The LCM of $1, x + 2, (x + 2)^2$ is, of course, $(x + 2)^2$, so
$f(x) = 1 \cdot \frac{(x + 2)^2}{(x + 2)^2} - \frac{3}{x + 2} \cdot \frac{x + 2}{x + 2}+ \frac{3}{(x + 2)^2}$

$f(x) = \frac{(x + 2)^2 - 3(x + 2) + 3}{(x + 2)^2}$

See if you can finish this from here.

-Dan
thank you SO much! i really appreciate it