determine real numbers a and b so that the expression 8sin^2(theta) + 2cos^2(theta) can be rewritten as (a)sin^2(theta) + b?
If you can help me out on how to start it would be great.
Use the identity:
sin²(x)+cos²(x)=1
In
8sin^2(theta) + 2cos^2(theta)
regroup to get:
6sin^2(theta) + 2(sin^2(theta)+cos^2(theta))
=6sin^2(theta) + 2
QED
To determine real numbers a and b such that the expression 8sin^2(theta) + 2cos^2(theta) can be rewritten as (a)sin^2(theta) + b, we need to use a trigonometric identity.
Starting with the given expression, let's use the identity sin^2(theta) + cos^2(theta) = 1, also known as the Pythagorean identity.
8sin^2(theta) + 2cos^2(theta)
= 8sin^2(theta) + 2(1 - sin^2(theta)) [using sin^2(theta) + cos^2(theta) = 1]
= 8sin^2(theta) + 2 - 2sin^2(theta)
= 6sin^2(theta) + 2
Now, we can rewrite the expression as (a)sin^2(theta) + b by factoring out the common factor of 2:
6sin^2(theta) + 2
= 2(3sin^2(theta) + 1)
So, we can see that a = 3 and b = 1 satisfy the condition for the given expression to be rewritten as (a)sin^2(theta) + b.
Therefore, the expression 8sin^2(theta) + 2cos^2(theta) can be rewritten as 3sin^2(theta) + 1.