# Math Help - ~Trigonometry

1. ## ~Trigonometry

Prove this identity:

$\frac{cos2A}{1+sin2A} = tan(\frac{\pi}{4} - A)$

Thanks I was having trouble proving this!
In the previous question I was able to prove:

$\frac{sin2A}{1 + cos2A} = tanA$

2. Originally Posted by slevvio
Prove this identity:

$\frac{cos2A}{1+sin2A} = tan(\frac{\pi}{2} - A)$

Thanks I was having trouble proving this!
I'm not surprised you are having trouble proving it. It isn't true.

-Dan

3. Oh dear, I copied the angle down wrong ! I have edited my post

4. Originally Posted by slevvio
Prove this identity:

$\frac{cos2A}{1+sin2A} = tan(\frac{\pi}{4} - A)$
That's better.

Let's do this one from the RHS since I don't know which form of $cos(2A)$ will be the most helpful.

$tan \left ( \frac{\pi}{4} - A \right ) = \frac{tan \left ( \frac{\pi}{4} \right ) - tan(A)}{1 + tan \left ( \frac{\pi}{4} \right ) \cdot tan(A)}$

$= \frac{1 - tan(A)}{1 + tan(A)}$

$= \frac{1 - \frac{sin(A)}{cos(A)}}{1 + \frac{sin(A)}{cos(A)}}$ <-- Clear the complex fraction

$= \frac{cos(A) - sin(A)}{cos(A) + sin(A)}$

Now I'm going to simplify by multiplying top and bottom by $cos(A) + sin(A)$:
$= \frac{cos^2(A) - sin^2(A)}{cos^2(A) + 2sin(A)cos(A) + sin^2(A)}$

$= \frac{cos(2A)}{1 + sin(2A)}$

Can I cook or can I cook?

-Dan

5. You certainly can sautée!