trigo (1)

• August 13th 2009, 03:17 AM
thereddevils
trigo (1)
Show for all values of a and b

$
\sin^2 a + \sin^2 (a+b) + \sin^2 (a+2b) = \frac{3}{2}-\frac{1}{2} \cos (2a+2b) (2 {\color{red}\cos b}+1)
$

This is what i did :

$
\frac{1}{2} [1 - \cos (2a)] + \frac{1}{2} [1 - \cos (2a+2b)] + \frac{1}{2} [1 - \cos(2a+4b)]
$

$=\frac{1}{2}[3-cos(2a+2b)-(cos2a+cos(2a+4b))]$

$=\frac{1}{2}[3-cos(2a+2b)-2cos(2a+2b)cos2b]$

$=\frac{1}{2}[3-cos(2a+2b)(1+2cos2b)]$

$
= \frac{3}{2} - \frac{1}{2} \cos (2a+2b) (2{\color{red}\cos (2b)} +1)
$

Why the difference in red ? Where did i go wrong ? Thanks for pointing that out .
• August 13th 2009, 03:24 AM
songoku
Hi thereddevils

I think you're right and the question is wrong :)