# How do you prove sin^(2)(x+π/4)-sin^(2)(x-π/4)=2sinx*cosx?

Feb 10, 2016

See proof.

#### Explanation:

We can prove the identity if we can start with the left-hand side and end up at the right side.

Along the way, let's use the following trigonometric identities:

 $\text{ } \sin \left(x + y\right) = \sin x \cos y + \cos x \sin y$

 $\text{ } \sin \left(x - y\right) = \sin x \cos y - \cos x \sin y$

and the binomial formula

 $\text{ } {\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

 $\text{ } {\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$

So, let's start:

${\sin}^{2} \left(x + \frac{\pi}{4}\right) - {\sin}^{2} \left(x - \frac{\pi}{4}\right) = {\left[\sin \left(x + \frac{\pi}{4}\right)\right]}^{2} - {\left[\sin \left(x - \frac{\pi}{4}\right)\right]}^{2}$

${\stackrel{\text{,")(" } =}{\sin \left(x\right) \cos \left(\frac{\pi}{4}\right) + \cos \left(x\right) \sin \left(\frac{\pi}{4}\right)}}^{2}$

$- {\left[\sin \left(x\right) \cos \left(\frac{\pi}{4}\right) - \cos \left(x\right) \sin \left(\frac{\pi}{4}\right)\right]}^{2}$

" " stackrel(",")(" "=) [sin^2x cos^2(pi/4) + 2 sinx cosx sin(pi/4)cos(pi/4) + cos^2x sin^2(pi/4)]

$- \left[{\sin}^{2} x {\cos}^{2} \left(\frac{\pi}{4}\right) - 2 \sin x \cos x \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right) + {\cos}^{2} x {\sin}^{2} \left(\frac{\pi}{4}\right)\right]$

$\text{ } = \textcolor{b l u e}{\cancel{{\sin}^{2} x {\cos}^{2} \left(\frac{\pi}{4}\right)}} + 2 \sin x \cos x \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right) + \textcolor{red}{\cancel{{\cos}^{2} x {\sin}^{2} \left(\frac{\pi}{4}\right)}}$

$\textcolor{b l u e}{\cancel{- {\sin}^{2} x {\cos}^{2} \left(\frac{\pi}{4}\right)}} + 2 \sin x \cos x \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right) \textcolor{red}{\cancel{- {\cos}^{2} x {\sin}^{2} \left(\frac{\pi}{4}\right)}}$

$= 4 \sin x \cos x \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)$

... remember that $\sin \left(\frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$...

$= 4 \sin x \cos x \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}$

$= \cancel{4} \sin x \cos x \cdot \frac{2}{\cancel{4}}$

$= 2 \sin x \cos x$

As we have safely arrived at the right-hand side, we haven proven the identity.

q.e.d.

I wasn't able to format the solution in a better way or find a way to have the whole term on one line during some steps. I hope that this is readable enough nevertheless.