How do you prove #sin (x+y)sin(x-y) = cos^2y-cos^2x#?

1 Answer
Lotusbluete
Jan 22, 2016

You should use the following trigonometrical identities:

#sin (x + y) = sin x cos y + cos x sin y #

#sin (x - y) = sin x cos y - cos x sin y #

This will lead you to:

#sin(x + y )sin (x - y)#

#= (sin x cos y + cos x sin y)(sin x cos y - cos x sin y)#

# = (color(blue)(sin x cos y) + color(red)(cos x sin y))(color(blue)(sin x cos y) - color(red)(cos x sin y))#

... use the formula #(a+b)(a-b) = a^2 - b^2#:

# = sin^2 x cos^2 y - cos^2 x sin^2 y#

... use #sin^2 x + cos^2 x = 1 " " <=> " " sin^2 x = 1 - cos^2 x#:

# = (1 - cos^2 x) cos^2 y - cos^2 x (1 - cos^2 y)#

# = cos^2 y - cos^2 x cos^2 y - cos^2 x + cos^2 x cos^2 y#

# = cos^2 y - cancel(cos^2 x cos^2 y) - cos^2 x + cancel(cos^2 x cos^2 y)#

# = cos^2 y - cos^2 x#

Related questions
See all questions in Sum and Difference Identities
Impact of this question
29608 views around the world
You can reuse this answer
Creative Commons License