Verify the function cos ß- sin² ß=1-2 sin² ß, starting point please.?

Question

Now, I dont' necessarily want a step by step explanation on how to do it, but I do wish I could have some help on how to start the problem.

Answer 1

cos^2b = 1- sin^2b (since cos^2b+sin^2b =1)

1- sin^2b -sin^2b = 1-2sin^2b

Answer 2

this is an identity that should read:

cos^2 B - sin^2 B = 1-2sin^2 B

start here: Cos^2 B = 1-sin^2 B
Then substitute into the original equation

OR you could return to the definition of the two functions

Sin B = y/r and Cos B = x/r
again, substitute the two values into the original identity to verify it

Answer 3

I think you mean cos^2 (B) - sin^2 (B).

Anyway, from the pythagorean identity sin^2 (B) + cos^ (B) = 1, we can solve for cos^2 (B) and then substitute it in for our expression, to get our desired result.

Answer 4

You need to use trig identities. For example, 1-2sin^2ß = 2cos^2ß - 1. cosß - sin^2ß = 2cos^2ß - 1; cosß - sin^2ß - cos^2ß = cos^2ß -1; Then sin^2ß+cos^2ß=1, so cosß = cos^2ß-1, etc

You can find a list of common identities here: http://en.wikipedia.org/wiki/Trigonometric_identity

Answer 5

cosA - sinA^2 = 1 - 2sinA^2

cosA = 1 - sinA^2

i believe you mean cosA^2 instead of cosA

http://www.math.com/tables/trig/identities.htm