Prove this identity:
$\displaystyle \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:
$\displaystyle \frac{sin2A}{1 + cos2A} = tanA $
Prove this identity:
$\displaystyle \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:
$\displaystyle \frac{sin2A}{1 + cos2A} = tanA $
That's better.
Let's do this one from the RHS since I don't know which form of $\displaystyle cos(2A)$ will be the most helpful.
$\displaystyle tan \left ( \frac{\pi}{4} - A \right ) = \frac{tan \left ( \frac{\pi}{4} \right ) - tan(A)}{1 + tan \left ( \frac{\pi}{4} \right ) \cdot tan(A)}$
$\displaystyle = \frac{1 - tan(A)}{1 + tan(A)}$
$\displaystyle = \frac{1 - \frac{sin(A)}{cos(A)}}{1 + \frac{sin(A)}{cos(A)}}$ <-- Clear the complex fraction
$\displaystyle = \frac{cos(A) - sin(A)}{cos(A) + sin(A)}$
Now I'm going to simplify by multiplying top and bottom by $\displaystyle cos(A) + sin(A)$:
$\displaystyle = \frac{cos^2(A) - sin^2(A)}{cos^2(A) + 2sin(A)cos(A) + sin^2(A)}$
$\displaystyle = \frac{cos(2A)}{1 + sin(2A)}$
Can I cook or can I cook?
-Dan