Asked by Andres

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.

Answers

Answered by Mgraph
8sin^2+2cos^2=8sin^2+2(1-sin^2)=>
a=6 & b=2
Answered by Anonymous
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
Answered by Anonymous
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) ]
Answered by Andres
thanks

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