Posted by Andres on Sunday, August 28, 2011 at 6:49pm.
determine real numbers a and b so that the expression 8sin^2(theta) + 2cos^2(theta) can be rewritten as asin^2(theta) + b
I just need help in how to start it.

Trigonometry  Mgraph, Sunday, August 28, 2011 at 7:11pm
8sin^2+2cos^2=8sin^2+2(1sin^2)=>
a=6 & b=2 
Trigonometry  Anonymous, Sunday, August 28, 2011 at 7:30pm
8 sin^2(theta) + 2 cos^2(theta) = 6 sin^2(theta) + 2 [ sin^2(theta) + 2 cos^2(theta) ] =
6 sin^2(theta) + 2 [ sin^2(theta) + cos^2(theta) ]
Remark:
sin^2(theta) + cos^2(theta) = 1
6 sin^2(theta) + 2 [ sin^2(theta) + cos^2(theta) ] = 6 sin^2(theta) + 2 * 1 =
6 sin^2(theta) + 2
8 sin^2(theta) + 2 cos^2(theta) = 6 sin^2(theta) + 2 =
6 sin^2(theta) + 2 = a sin^2(theta) + b
Obviously:
a = 6
b = 2 
Trigonometry  Anonymous, Sunday, August 28, 2011 at 7:32pm
8 sin^2(theta) + 2 cos^2(theta) = 6 sin^2(theta) + 2 sin^2(theta) + 2 cos^2(theta) = 6 sin^2(theta) + 2 [ sin^2(theta) + cos^2(theta) ]

Trigonometry  Andres, Sunday, August 28, 2011 at 8:12pm
thanks