# Prove that (cosx-sinx)^2 = 1 – sin2x?

additionally, is it possible to offer any hints/easy tips to consider how exactly to do any trig identities, thanks a lot.

Note : sin^2 p + cos^2 p = 1
sin 2p = sin ( p + p) = sin p cos p + cos p sin p = 2 sin p cos p
(cosx-sinx)^2
=cos^2 x -2sinx cosx + sin^2 x
=sin^2 x + cos^2 x – 2sinx cosx
= 1 – sin2x

cos^2x – 2cosxsinx + sin^2x = 1 – sin2x
1 – 2cosxsinx = 1 – sin2x
1 – sin2x = 1 – sin2x
An over-all move to make should seek out Identities and Formulas which can be inside issue. As an example, within issue, a great place to begin could be broadening the left-hand Side. After that, you can view that there’s a Pythagorean identification; sin^2x + cos^2x = 1, making demonstrating simpler.
And yes it really helps to view what you ought to reach the conclusion. If you’d like a-1 or a sin2x, seek out their particular equivalences inside various Identities/Proofs.

1 Sin2x Formula

Increase the remaining part getting:
sin^2(x)+cos^2(x)-2 sin(x) cos(x) = 1-sin(2 x)
We realize that sin^2(x)+cos^2(x) = 1, to help you replace that directly into get:
1-2 sin(x) cos(x) = 1-sin(2 x)
Today, we additionally understand that sin(x) * cos(x) = 1/2sin(2x)
Stick this in therefore the 1/2 will terminate utilizing the 2 getting:
1-sin(2 x) = 1-sin(2 x)
In terms of trig identification tips, simply keep a listing of the fundamental identities useful and constantly you will need to streamline around feasible.

(cosx-sinx)^2 = cos^2^x – 2cosxsinx + sin^2^x
cos^2^x + sin^2^x = 1
– 2cosxsinx = – sin2x
==> (cosx-sinx)^2 = 1 – sin2x