How do you prove #sin^(2)(x+π/4)-sin^(2)(x-π/4)=2sinx*cosx#?
1 Answer
See proof.
Explanation:
We can prove the identity if we can start with the left-hand side and end up at the right side.
Along the way, let's use the following trigonometric identities:
[1]
#" " sin(x+y) = sinx cos y + cos x sin y# [2]
#" " sin(x-y) = sinx cos y - cos x sin y#
and the binomial formula
[3]
#" "(a + b)^2 = a^2 + 2ab + b^2# [4]
#" "(a - b)^2 = a^2 - 2ab + b^2#
So, let's start:
# stackrel("[1],[2]")(" "=) [sin(x) cos(pi/4) + cos(x) sin(pi/4)]^2#
#- [sin(x) cos(pi/4) - cos(x) sin(pi/4)]^2#
#- [sin^2 x cos^2(pi/4) - 2sinx cosx sin(pi/4)cos(pi/4) + cos^2 x sin^2(pi/4)]#
#color(blue)(cancel(- sin^2 x cos^2(pi/4))) + 2sinx cosx sin(pi/4)cos(pi/4) color(red)(cancel(- cos^2 x sin^2(pi/4)))#
# = 4 sin x cos x sin(pi/4)cos(pi/4)#
... remember that
# = 4sin x cos x * sqrt(2)/2*sqrt(2)/2#
# = cancel(4)sin x cos x * 2/cancel(4)#
# = 2sin x cos x #
As we have safely arrived at the right-hand side, we haven proven the identity.
q.e.d.
I wasn't able to format the solution in a better way or find a way to have the whole term on one line during some steps. I hope that this is readable enough nevertheless.