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.
8sin^2+2cos^2=8sin^2+2(1-sin^2)=>
a=6 & b=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
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) ]
thanks
To rewrite the given expression, 8sin^2(theta) + 2cos^2(theta), as asin^2(theta) + b, we need to make use of the identities of trigonometric functions. One such identity that will be helpful in solving this problem is:
sin^2(theta) + cos^2(theta) = 1
Using this identity, we can rewrite 8sin^2(theta) + 2cos^2(theta) as:
8sin^2(theta) + 2cos^2(theta) = 8sin^2(theta) + 2(1 - sin^2(theta))
Now, simplify the expression:
8sin^2(theta) + 2(1 - sin^2(theta)) = 8sin^2(theta) + 2 - 2sin^2(theta)
Combining like terms:
8sin^2(theta) + 2 - 2sin^2(theta) = (8 - 2)sin^2(theta) + 2
Simplifying further:
(8 - 2)sin^2(theta) + 2 = 6sin^2(theta) + 2
Now we need to rewrite this expression in the form asin^2(theta) + b. To do so, we can factor out 2 from the terms:
6sin^2(theta) + 2 = 2(3sin^2(theta) + 1)
Comparing this to the desired form asin^2(theta) + b:
asin^2(theta) + b = 2(3sin^2(theta) + 1)
From the comparison, we can deduce that a = 2 and b = 2.
So, the expression 8sin^2(theta) + 2cos^2(theta) can be rewritten as 2sin^2(theta) + 2.