Melody

That is really cute - thanks anon :)

Start by factoring out the 5 :)

180(8-2) = 180*6

The coetfficent is the number in front of the letter/s

For example:

$$5x^2+6x-3-12xy+xy^2$$

the coefficient of x^2 is 5

the coefficient of x is 6

the coefficient of xy is -12

the coefficient of xy^2 is 1

and

the constant is -3

I'll use the calc and see.

$${\left(-{\mathtt{1}}\right)}^{{\mathtt{2}}} = {\mathtt{1}}$$

$${\mathtt{\,-\,}}\left({{\mathtt{1}}}^{{\mathtt{2}}}\right) = -{\mathtt{1}}$$        The calc put the brackets in - I did not.  But the answer is correct.  :)

Neither of us made the site but thank you.  

This site is is owned and built by a brilliant man by the name of Mr. Massow.  :))

It could be any length between 0 and 14.75 not inclusive.

It would depend on the size of the angle between the other two sides :/

$$\\\frac{2^3*2^0}{2}-8*2^{-2}\\\\
=\frac{8*1}{2}-8*\frac{1}{2^{+2}}\\\\
=\frac{4*1}{1}-8*\frac{1}{4}\\\\
=4-2*\frac{1}{1}\\\\
=4-2\\\\
=2$$

$$\\LHS
=\frac{( ((-5x-3)/(2x-1))-3 )}{( 2((-5x-3)/(2x-1)) + 5)}\\\\
=\left(\frac{(-5x-3)}{(2x-1)}-3 \right )\div \left( 2\times\frac{(-5x-3)}{(2x-1)} + 5\right)\\\\
=\left(\frac{(-5x-3)-3(2x-1)}{(2x-1)} \right )\div \left(\frac{2(-5x-3)+5(2x-1)}{(2x-1)} \right)\\\\
=\left(\frac{(-5x-3)-3(2x-1)}{(2x-1)} \right )\times \left(\frac{(2x-1)}{2(-5x-3)+5(2x-1)} \right)\\\\
=\left(\frac{-5x-3-6x+3}{1} \right )\times \left(\frac{1}{-10x-6+10x-5} \right)\\\\
=\left(\frac{-11x}{1} \right )\times \left(\frac{1}{-11} \right)\\\\
=x\\\\
=RHS \qquad QED$$

Answered here 

http://web2.0calc.com/questions/prove-sin4x-sin2x-cos4x-cos2x