# How to establish this identity?

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

#= cos x - sin x#

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**See the steps below!**

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

This means,

Manipulate the left hand side:

Apply the difference of squares rule

So that we have:

Take negative common in the numerator to get:

Canel out the common factor

Simplify:

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

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See below.

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Cancel out

Proved.

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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

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