# Math Help - Trig ratio help

1. ## Trig ratio help

I feel terrible for making yet another thread, but I've been stuck on this for 30 mins:

LS=cos(2x)/(1-sin2x)

RS=(cosx+sinx)/(cosx-sinx)

2. Since you see double angles on the LS, convert them.

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

$sin(2x) = 2 \sin(x)\cos(x)$

It should be okay then

3. Originally Posted by Unknown008
Since you see double angles on the LS, convert them.

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

$sin(2x) = 2 \sin(x)\cos(x)$

It should be okay then
I don't suppose you could continue it. This is the only thing that's giving me trouble and I'm still kind of lost

4. Hello, youngb11!

Another approach . . .

$\dfrac{\cos 2x}{1-\sin2x} \;=\;\dfrac{\cos x+\sin x}{\cos x-\sin x}$

Multiply the right side by $\frac{\cos x - \sin x}{\cos x - \sin x}$

. . $\displaystyle \frac{\cos x + \sin x}{\cos x - \sin x}\cdot\frac{\cos x - \sin x}{\cos x - \sin x} \;=\;\frac{\cos^2\!x - \sin^2\!x}{\cos^2\!x - 2\cos x\sin x + \sin^2\!x}$

. . . . $\displaystyle =\;\frac{\overbrace{\cos^2\!x - \sin^2\!x}^{\text{This is }\cos2x}}{\underbrace{(\sin^2\!x + \cos^2\!x)}_{\text{This is 1}} - \underbrace{2\sin x\cos x}_{\text{This is }\sin2x}} \;=\;\frac{\cos2x}{1-\sin2x}$

5. Thanks a lot! I never saw that approach and was intent on simplifying the LS to equal the RS. Thanks again!

6. Just use the identities I gave you otherwise.

$\dfrac{\cos 2x}{1-\sin2x} = \dfrac{\cos^2(x) - \sin^2(x)}{1 - 2\cos(x)\sin(x)}$

Here, the trick is to spot that the numerator is a difference of two squares and that the numerator can be a square by using the Pythagorean identity, $\cos^2(x) + \sin^2(x) = 1$

This said, you get:

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

You then factor the denominator;

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

Some things cancel out, leaving the RS.

You can also see that it's the reverse process that Soroban posted