1. ## Trigo identities

hey, ive got 2 questions here in which i dont know how to do. Im completely clueless for question 1 but i did some for question 2. hopefully someone can help me ^^ thanks!

Q1) Prove $\displaystyle (1 - tan A)(tan 2A + Sec 2A) = 1 + tan A$
Q2) Prove $\displaystyle cos^4 A = \frac {1}{8} (3 + 4cos 2A + cos 4A)$
for this, i got
$\displaystyle \frac {3}{8} + cos^2 A - \frac{1}{2} + \frac{1}{8} cos 2A cos 2A - \frac{1}{8} sin 2A sin 2A$
=
$\displaystyle cos^2 A - \frac {1}{8} + \frac {1}{8} ( 4cos^4 A - 4cos^2 A + 1) - \frac {1}{2} sin^2 A cos^2 A$
=
$\displaystyle cos^2 A - \frac {1}{8} + \frac {1}{2} cos^4 A - \frac{1}{2} cos^2 A + \frac{1}{8} - \frac {1}{2} sin^2 A cos^2 A$
=
$\displaystyle 2cos^2 A - \frac {1}{4} + cos^4 A - cos^2 A + \frac {1}{4} - sin^2 A cos^2 A$
=
$\displaystyle cos^2 A - sin^2 A cos^2 A - \frac {1}{4} + \frac {1}{4} + cos^4 A$
and i knew that it was from this step that something is wrong, but i dont know what it is ><

2. Originally Posted by teddybear67
$\displaystyle cos^2 A - \frac {1}{8} + \frac {1}{2} cos^4 A - \frac{1}{2} cos^2 A + \frac{1}{8} - \frac {1}{2} sin^2 A cos^2 A$
=
$\displaystyle 2cos^2 A - \frac {1}{4} + cos^4 A - cos^2 A + \frac {1}{4} - sin^2 A cos^2 A$
I do not understand why you multiplied everything by 2 ?

You are close to the end :

$\displaystyle \cos^2 A - \frac {1}{8} + \frac {1}{2} \cos^4 A - \frac{1}{2} \cos^2 A + \frac{1}{8} - \frac {1}{2} \sin^2 A \cos^2 A$ = $\displaystyle \frac{1}{2} \cos^2 A + \frac {1}{2} \cos^4 A - \frac {1}{2} \1-\cos^2 A) \cos^2 A = \cos^4 A$