How do you verify #(sin B + cos B)(sin B - cos B)= 2 sin^2B - 1#?

2 Answers
Gió
Aug 3, 2015

Have a look:

Explanation:

We can multiply the brackets and get:
#sin^2Bcancel(-sinBcosB)cancel(+sinBcosB)-cos^2B=2sin^2B-1#
#sin^2B-cos^2B=2sin^2B-1#
But: #sin^2B+cos^2B=1#
So:
#sin^2B-cos^2B=2sin^2B-sin^2B-cos^2B#
And:
#sin^2B-2sin^2B+sin^2B=cos^2B-cos^2B#
#0=0#

Nghi N.
Aug 3, 2015

Verify # (sin B - cos B)(sin B + cos B) = 2sin^2 B - 1#

Explanation:

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

#- (cos^2B - sin^2 B) = - (1 - 2sin^2 B) #

#- cos (2B) = - cos (2B)#

Reminder: #cos 2a = 1 - 2sin^2a = cos^2 a - sin^2 a#

Related questions
See all questions in Proving Identities
Impact of this question
3064 views around the world
You can reuse this answer
Creative Commons License