determine real numbers a and b so that the expression 8sin^2(theta) + 2cos^2(theta) can be rewritten as asin^2(theta) + b

Question

I just need help in how to start it.

Answer 1

8sin^2+2cos^2=8sin^2+2(1-sin^2)=>
a=6 & b=2

Answer 2

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

Answer 3

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) ]

Answer 4

thanks