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(1-sin^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

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