Come si verifica # cos ^ 2 (x) -sin ^ 2 (x) = 1-2sin ^ 2 (x) #?

RTP: #cos^2x-sin^2x=1-2sin^2x#

LHS:
#cos^2x-sin^2x#

Richiamare: #cos^2x+sin^2x=1# so #cos^2x=1-sin^2x#

= #(1-sin^2x)-sin^2x#

=#1-sin^2x-sin^2x#

= #1-2sin^2x#

= RHS

Perciò: #cos^2x-sin^2x=1-2sin^2x#