Question

How to establish this identity?

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

Answer 1

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

Answer 2

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.

Answer 3

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)`

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`\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)}`

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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)}`

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Take negative common in the numerator to get:

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

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

`-(\sin (x)-\cos (x))`

Simplify:

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

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