Prove that:
cos^2A - sin^2A = 1 - 2sin^2A
In words...
cos squared A - sin squared A = one minus two sin squared A
2007-05-29 01:16:29 · 4 answers · asked by stacey_b 2 in Science & Mathematics ➔ Mathematics
Thanks Robert.. yours was really easy to understand!! Thanks again!!
2007-05-29 01:37:04 · update #1
use: cos(2A) = 1 - 2sin2(A)
and rewrite: cos^2A - sin^2A
as: 1 - 2sin2(A) - sin^2A
simplify: 1 - 2sin^2A
2007-05-29 01:19:10 · answer #1 · answered by mikedotcom 5 · 0⤊ 1⤋
Since all sin and cos functions are of A, I'll omit the A for simplicity.
cos^2 - sin^2 = 1 - 2 * sin^2
Add 2 * sin^2 to both sides:
cos^2 + (-sin^2 + 2 * sin^2) = 1 + (-2 * sin^2 + 2 * sin^2)
cos^2 + sin^2 = 1
This should be adequate. If not, then prove the identity above by using the Pythagorean theorem, if necessary.
2007-05-29 08:28:23 · answer #2 · answered by Carl M 3 · 0⤊ 0⤋
cos² x = 1 - sin² x (identity)
cos²A - sin²A = 1 - 2sin²A
(1 - sin² A) - sin² A = 1 - 2sin²A
1 - 2sin² A = 1 - 2sin² A
.
2007-05-29 08:33:55 · answer #3 · answered by Robert L 7 · 0⤊ 0⤋
ohk.
cos^2A - sin^2A
1- sin^2A - sin^2A [sin^2A +cos^2A=1]
1 - 2sin^2A
2007-05-29 08:19:29 · answer #4 · answered by jedi Knight 2 · 0⤊ 0⤋