How do you rewrite (sin x - cos x)(sin x + cos x)? Thank you.

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Answer 1 · 2018-04-06T00:53:56

$sin^2x-cos^2x$

You're probably used to dealing with this only in quadratics, but the expression is in the difference of squares pattern

$(a-b)(a+b)=a^2-b^2$

where $color(red)(a=sinx)$ and $color(blue)(b=cosx)$

We can just plug in our values for $a$ and $b$ into our difference of squares expression, and we get:

$color(red)((sinx))^2-color(blue)((cosx))^2$

Which can be rewritten as

$sin^2x-cos^2x$

If you don't believe me, we can FOIL this expression to make sure:

With FOIL, we multiply the first, outside, inside and last terms and add the result. Thus, we have:

Now we have

$color(red)((sinx))^2+cancel(sinxcosx-sinxcosx)-color(blue)((cosx))^2$

The middle terms obviously cancel out, and we can rewrite this as

$sin^2x-cos^2x$

The key realization is that the original expression in question was in a difference of squares pattern.

If you have a $(a-b)(a+b)$ pattern, you can always use the difference of squares identity to quickly simplify it.

Hope this helps!

Answer 2 · 2018-04-06T02:58:27

$- cos 2x$

$f(x) = (sin x - cos x)(sin x + cos x) = (sin^2 x - cos^2 x)$
Reminder of trig identity:
$cos 2x = cos^2 x - sin^2 x$
Finally,
$f(x) = - cos 2x$