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How to establish this identity?

(sin^2(-x)-cos^2(-x))/(sin(-x)-cos(-x)  $= \cos x - \sin x$

Feb 22, 2018

We seek to prove the identity:

$\frac{{\sin}^{2} \left(- x\right) - {\cos}^{2} \left(- x\right)}{\sin \left(- x\right) - \cos \left(- x\right)} \equiv \cos x - \sin x$

Consider the LHS:

$L H S = \frac{{\sin}^{2} \left(- x\right) - {\cos}^{2} \left(- x\right)}{\sin \left(- x\right) - \cos \left(- x\right)}$

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

$\sin \left(- x\right) = - \sin \left(x\right)$ and $\cos \left(- x\right) = \cos \left(x\right)$

Then:

$L H S = \frac{{\sin}^{2} \left(x\right) - {\cos}^{2} \left(x\right)}{- \sin x - \cos x}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\left(\sin x + \cos x\right) \left(\sin x - \cos x\right)}{- \left(\sin x + \cos x\right)}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - \left(\sin x - \cos x\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \cos x - \sin x \setminus \setminus \setminus$ QED

Feb 22, 2018

See below.

Explanation:

We have $\frac{{\sin}^{2} \left(- x\right) - {\cos}^{2} \left(- x\right)}{\sin \left(- x\right) - \cos \left(- x\right)}$

Since $\sin \left(- x\right) = - \sin \left(x\right)$ and $\cos \left(- x\right) = \cos \left(x\right)$, we can simplify this to:

$\frac{{\left(- \sin \left(x\right)\right)}^{2} - {\cos}^{2} \left(x\right)}{- \cos \left(x\right) - \sin \left(x\right)}$

Since ${\left(- a\right)}^{2} = {a}^{2}$, we write the above as:

$\frac{{\sin}^{2} \left(x\right) - {\cos}^{2} \left(x\right)}{- 1 \left(\cos \left(x\right) + \sin \left(x\right)\right)}$

Since ${x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)$, we can write:

$\frac{\left(\sin \left(x\right) + \cos \left(x\right)\right) \left(\sin \left(x\right) - \cos \left(x\right)\right)}{- 1 \left(\cos \left(x\right) + \sin \left(x\right)\right)}$

Cancel out $\sin \left(x\right) + \cos \left(x\right)$:

$\frac{\sin \left(x\right) - \cos \left(x\right)}{-} 1$

$\cos \left(x\right) - \sin \left(x\right)$

Proved.

Feb 22, 2018

See the steps below!

Explanation:

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

This means, $\sin \left(- x\right) = - \sin \left(x\right)$ $\setminus \setminus \setminus$ and $\setminus \setminus \setminus$ $\cos \left(- x\right) = \cos \left(x\right)$





$\setminus \frac{\setminus {\sin}^{2} \left(- x\right) - \setminus {\cos}^{2} \left(- x\right)}{\setminus \sin \left(- x\right) - \setminus \cos \left(- x\right)} = \setminus \cos \left(x\right) - \setminus \sin \left(x\right)$

Manipulate the left hand side:

$= \setminus \frac{{\left(- \setminus \sin \left(x\right)\right)}^{2} - \setminus {\cos}^{2} \left(x\right)}{- \setminus \cos \left(x\right) - \setminus \sin \left(x\right)}$





Apply the difference of squares rule ${x}^{2} - {y}^{2} = \left(x + y\right) \left(x - y\right)$ in the numerator:

$\setminus {\sin}^{2} \left(x\right) - \setminus {\cos}^{2} \left(x\right) = \left(\setminus \sin \left(x\right) + \setminus \cos \left(x\right)\right) \left(\setminus \sin \left(x\right) - \setminus \cos \left(x\right)\right)$

So that we have:

$= \setminus \frac{\left(\setminus \sin \left(x\right) + \setminus \cos \left(x\right)\right) \left(\setminus \sin \left(x\right) - \setminus \cos \left(x\right)\right)}{- \setminus \cos \left(x\right) - \setminus \sin \left(x\right)}$




Take negative common in the numerator to get:

$= \setminus \frac{\left(\setminus \sin \left(x\right) + \setminus \cos \left(x\right)\right) \left(\setminus \sin \left(x\right) - \setminus \cos \left(x\right)\right)}{- \left(\setminus \cos \left(x\right) + \setminus \sin \left(x\right)\right)}$




Canel out the common factor $\setminus \setminus \setminus$ $\setminus \cos \left(x\right) + \setminus \sin \left(x\right)$ to get:

$- \left(\setminus \sin \left(x\right) - \setminus \cos \left(x\right)\right)$

Simplify:

$= \setminus \cos \left(x\right) - \setminus \sin \left(x\right)$




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