# Math Help - Simplifying Trig Equation

1. ## Simplifying Trig Equation

Hi can anyone show me step by step simplification of this equation

$((A+C)/2)^2 - (((A-C)/2)cos2x + Bsin2x)^2 -(Bcos2x-((A-C)/2)sin2x)^2$

I know its suppose to come out to equal $AC-B^2$

Thank you so much in advance... I can't seem to get the write answer.

2. Hello, mikec5!

Simplify: . $\left(\frac{A+C}{2}\right)^2 - \bigg[\left(\frac{A-C}{2}\right)\cos2x + B\sin2x\bigg]^2 -\bigg[B\cos2x-\left(\frac{A-C}{2}\right)\sin2x\bigg]^2$

Answer: . $AC-B^2$
We have:

$\left(\frac{A+C}{2}\right)^2 -\bigg[\left(\frac{A-C}{2}\right)^2\cos^2\!2x + B(A-C)\sin2x\cos2x + B^2\sin^2\!2x\bigg]$

. . . . $-\bigg[B^2\cos^2\!2x - B(A-C)\sin2x\cos2x + \left(\frac{A-C}{2}\right)^2\sin^2\!2x\bigg]$

$=\;\left(\frac{A+C}{2}\right)^2 - \left(\frac{A-C}{2}\right)^2\cos^2\!2x - {\color{red}\rlap{||||||||||}}B(A-C)\sin2x\cos2x - B^2\sin^2\!2x$

. . . . $-B^2\cos^2\!2x + {\color{red}\rlap{||||||||||}}B(A-C)\sin2x\cos2x - \left(\frac{A-C}{2}\right)^2\sin^2\!2x
$

Re-arrange the terms:

. . $\left(\frac{A+C}{2}\right)^2 -\;\left(\frac{A-C}{2}\right)^2\!\cos^2\!2x \;- \;\left(\frac{A-C}{2}\right)^2\!\sin^2\!2x \;-\; B^2\sin^2\!2x \;-\; B^2\cos^2\!2x$

Factor:

. . $\left(\frac{A+C}{2}\right)^2 - \left(\frac{A-C}{2}\right)^2\!\underbrace{\left(\cos^2\!2x + \sin^2\!2x\right)}_{\text{This is 1}} \;-\; B^2\underbrace{\left(\sin^2\!2x + \cos^2\!2x\right)}_{\text{This is 1}}$

. . $= \;\left(\frac{A+C}{2}\right)^2 - \left(\frac{A-C}{2}\right)^2 - B^2$

. . $=\;\frac{A^2 + 2AC + C^2}{4} - \frac{A^2 - 2AC + C^2}{4} - B^2$

. . $= \;\frac{4AC}{4} - B^2$

. . $=\;AC - B^2$