• Oct 30th 2009, 08:02 AM
Android

2Sin^2(x+pi/4) = 1 + Sin2x

I was thinking that I have to include cofunctions somehow, but have no idea how! This has been bugging me for too long! (Headbang)

Thanks!
Andrew
• Oct 30th 2009, 08:21 AM
ramiee2010
Quote:

Originally Posted by Android

2Sin^2(x+pi/4) = 1 + Sin2x

I was thinking that I have to include cofunctions somehow, but have no idea how! This has been bugging me for too long! (Headbang)

Thanks!
Andrew

use identity
$\cos {2x}= 1-2 \sin^2 {x}$
if stuck look spoiler
Spoiler:

$\cos {2x}= 1-2 \sin^2 {x} \quad \Rightarrow \quad \boxed {2 \sin^2 {x} = 1-\cos {2x}}$
$\therefore LHS=2Sin^2(x+pi/4)=1-\cos { \left ( 2 \cdot (x+ \frac {pi}{4}) \right ) }$
$=1- \cos { \left ( 2x+\frac{\pi}{2} \right )}=1+\sin {2x} =RHS$
• Oct 30th 2009, 08:39 AM
Android
That all makes perfect sense! Thank you SO much!