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How to establish this identity?

#(sin^2(-x)-cos^2(-x))/(sin(-x)-cos(-x) #
#= cos x - sin x#

3 Answers
Feb 22, 2018

We seek to prove the identity:

# (sin^2(-x)-cos^2(-x))/(sin(-x)-cos(-x)) -= cos x - sin x #

Consider the LHS:

# LHS = (sin^2(-x)-cos^2(-x))/(sin(-x)-cos(-x)) #

Using the fact that sine is and function and cosine is an even function we have:

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

Then:

# LHS = (sin^2(x)-cos^2(x))/(-sinx-cosx) #

# \ \ \ \ \ \ \ \ = ((sinx+cosx)(sinx-cosx))/(-(sinx+cosx)) #

# \ \ \ \ \ \ \ \ = -(sinx-cosx) #

# \ \ \ \ \ \ \ \ = cosx-sinx \ \ \ # QED

Feb 22, 2018

Answer:

See below.

Explanation:

We have #(sin^2(-x)-cos^2(-x))/(sin(-x)-cos(-x))#

Since #sin(-x)=-sin(x)# and #cos(-x)=cos(x)#, we can simplify this to:

#((-sin(x))^2-cos^2(x))/(-cos(x)-sin(x))#

Since #(-a)^2=a^2#, we write the above as:

#(sin^2(x)-cos^2(x))/(-1(cos(x)+sin(x)))#

Since #x^2-y^2=(x+y)(x-y)#, we can write:

#((sin(x)+cos(x))(sin(x)-cos(x)))/(-1(cos(x)+sin(x)))#

Cancel out #sin(x)+cos(x)#:

#(sin(x)-cos(x))/-1#

#cos(x)-sin(x)#

Proved.

Feb 22, 2018

Answer:

See the steps below!

Explanation:

Remember that, #sin# is an odd function while #cos# is an even function .

This means, #sin(-x)=-sin(x)# #\ \ \ # and #\ \ \ # #cos(-x)=cos(x)#

# #
# #
# #

#\frac{\sin ^2(-x)-\cos ^2(-x)}{\sin (-x)-\cos (-x)}=\cos (x)-\sin (x)#

Manipulate the left hand side:

#=\frac{(-\sin (x))^2-\cos ^2(x)}{-\cos (x)-\sin (x)}#

# #
# #
# #

Apply the difference of squares rule #x^2-y^2=(x+y)(x-y)# in the numerator:

#\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)-\cos (x))#

So that we have:

#=\frac{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{-\cos (x)-\sin (x)}#

# #
# #

Take negative common in the numerator to get:

#=\frac{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{-(\cos (x)+\sin (x))}#

# #
# #

Canel out the common factor #\ \ \ # #\cos (x)+\sin (x)# to get:

#-(\sin (x)-\cos (x))#

Simplify:

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

# #
# #

That's it. We just showed that the two sides could take the same form!