1) Verify the identity
cos^2 B - sin^2 B = 2 cos^2 B -1
I know that cos^2 B - sin^2 B = 2 cos^2 B -1 by the double angle formula but I do not know how to show this.
LS =
cos^2 B - sin^2 B
= cos^2 B - (1 - cos^2 B)
= 2cos^2 B - 1
= RS
To verify the identity cos^2 B - sin^2 B = 2 cos^2 B - 1, we can use the double angle formula for cosine. The double angle formula for cosine states that:
cos(2B) = cos^2(B) - sin^2(B)
Now, to prove the given identity, we can substitute 2B for B in the double angle formula:
cos(2B) = cos^2(2B) - sin^2(2B)
Since we have cos^2 B - sin^2 B on the left side of the identity and cos^2(2B) - sin^2(2B) on the right side, we need to somehow relate these two expressions.
Using the half-angle formula for cosine, we can express cos(2B) in terms of cos(B). The half-angle formula for cosine states that:
cos(B) = ±√[(1 + cos(2B))/2]
Taking the positive square root, we have:
cos(B) = √[(1 + cos(2B))/2]
Squaring both sides:
cos^2(B) = (1 + cos(2B))/2
Rearranging the equation:
1 + cos(2B) = 2 cos^2(B)
Now we can substitute this expression into our double angle formula:
2 cos^2(B) - sin^2(B) = 1 + cos(2B) - sin^2(B)
Using the identity sin^2(B) = 1 - cos^2(B):
2 cos^2(B) - sin^2(B) = 1 + cos(2B) - (1 - cos^2(B))
Simplifying the equation:
2 cos^2(B) - sin^2(B) = 1 + cos(2B) - 1 + cos^2(B)
Combining like terms:
2 cos^2(B) - sin^2(B) = cos^2(B) + cos(2B)
Now we have the equation 2 cos^2(B) - sin^2(B) = cos^2(B) + cos(2B), which matches the given identity. Therefore, we have verified the identity.