2. Originally Posted by Macleef

1) Prove:

sin2x / 1 - cos2x = cotx

My Attempt:
LS:
= 2sinxcosx / - 1 - (1 - 2sin^2x)
= 2sinxcosx / - 1 + 2sin^2x
= cosx / sinx - 1

Is there a typo? -1 - 1 = -2, not -1. And if you can't factor a 2 out of everything, then you can't cancel the 2, so the last line here is wrong. The same goes for sin(x).

$\frac{sin(2x)}{1 - cos(2x)} = cot(x)$

$\frac{2sin(x)cos(x)}{1 - cos(2x)} = \frac{cos(x)}{sin(x)}$

$\frac{2sin(x)cos(x)}{1 - (1 - 2sin^2(x)} = \frac{cos(x)}{sin(x)}$

$\frac{2sin(x)cos(x)}{1 - 1 + 2sin^2(x)} = \frac{cos(x)}{sin(x)}$

$\frac{2sin(x)cos(x)}{2sin^2(x)} = \frac{cos(x)}{sin(x)}$

$\frac{cos(x)}{sin(x)} = \frac{cos(x)}{sin(x)}$

You had the right idea, but you didn't have the details right.

-Dan

3. Thanks I see where I went wrong now, can you also please help with the rest?

Edit:

I don't know how to use the parenthesis?

4. Originally Posted by Macleef
I don't know how to use the parenthesis?
The kid is crying so I can't.

You don't know how to use parenthesis??? I'm talking about the difference between
$a + b/c + d = a + \frac{b}{c} + d$
vs.
$(a + b)/(c + d) = \frac{a + b}{c + d}$

You use them to group terms together.

-Dan

5. Originally Posted by topsquark
The kid is crying so I can't.

You don't know how to use parenthesis??? I'm talking about the difference between
$a + b/c + d = a + \frac{b}{c} + d$
vs.
$(a + b)/(c + d) = \frac{a + b}{c + d}$

You use them to group terms together.

-Dan
What? The kid is crying so you can't????

I can group them...but not with the latex feature because I don't know how to use it

6. Hello, Macleef!

4) Prove: . $8\cos^4\!x \:= \:\cos4x + 4\cos2x + 3$
We'll use the identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}$ . . . twice.

We have: . $8\left(\cos^2\!x\right)^2 \;=\;8\left(\frac{1 + \cos2x}{2}\right)^2 \;=\;\;8\left(\frac{1 + 2\cos2x + \cos^2\!2x}{4}\right)$

. . $= \;\;2\left(1 + 2\cos2x + \cos^2\!2x\right) \;\;=\;\;2\left(1 + 2\cos2x + \frac{1 + \cos4x}{2}\right)$

. . $= \;\;2 + 4\cos2x + 1 + \cos4x \;\;=\;\;{\color{blue}\cos4x + 4\cos2x + 3}$

7. Originally Posted by Soroban
Hello, Macleef!

We'll use the identity: . $\cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}$ . . . twice.

We have: . $8\left(\cos^2\!x\right)^2 \;=\;8\left(\frac{1 + \cos2x}{2}\right)^2 \;=\;\;8\left(\frac{1 + 2\cos2x + \cos^2\!2x}{4}\right)$

. . $= \;\;2\left(1 + 2\cos2x + \cos^2\!2x\right) \;\;=\;\;2\left(1 + 2\cos2x + \frac{1 + \cos4x}{2}\right)$

. . $= \;\;2 + 4\cos2x + 1 + \cos4x \;\;=\;\;{\color{blue}\cos4x + 4\cos2x + 3}$

wow, you lost me -_- from the beginning when you wrote, "we'll use this identity"

8. Originally Posted by Macleef
wow, you lost me -_- from the beginning when you wrote, "we'll use this identity"
There are two ways to get at this identity. I'll use the one you will have been likely to see:
$cos(2 \theta) = 2cos^2(\theta) - 1$

$2cos^2(\theta) = cos(2\theta) + 1$

$cos^2(\theta) = \frac{cos(2\theta) + 1}{2}$

(The quick way to do this is using a half-angle identity.)

-Dan

9. Originally Posted by Macleef