1. ## hard 2

2. There has to be something more behind this. What chapter of what book is it taken from, or what part of what course are you working with right now?

-Dan

3. $a^2-b^2+ac=\sin ^25-\sin ^249+\sin 5\sin 87=$
$=(\sin 5-\sin 49)(\sin 5+\sin 49)+\sin 5\sin 87=$
$=-2\sin 22\cos 27\cdot 2\sin 27\cos 22+\sin 5\sin 87=$
$=-\sin 44\sin 54+\sin 5\sin 87=-\frac{1}{2}(\cos 10-\cos 98-\cos 82+\cos 92)=$
$=-\frac{1}{2}(2\cos 51\cos 41-2\cos 90\cos 5)=-\cos 41\cos 51$

Now the equality becomes
$\sin 73=\frac{\cos 41\cos 51}{4\sin 5\cos 51\cos 41+\sin 5-\sin 49+\sin 87}$

We have $4\sin 5\cos 51\cos 41=2(\sin 56-\sin 46)\cos 41=2\sin 56\cos 41-2\sin 46\cos 41=$
$=\sin 15+\sin 97-\sin 5-\sin 87$
The equality becomes
$\sin 73=\frac{\cos 41\cos 51}{\sin 15+\sin 97-\sin 49}\Leftrightarrow$
$\Leftrightarrow \cos 41\cos 51=\sin 15\sin 73+\sin 97\sin 73-\sin 49\sin 73\Leftrightarrow$
$\Leftrightarrow \cos 92+\cos 10=\cos 58-\cos 88+\cos 24-\cos 170-\cos 24+\cos 122\Leftrightarrow$
$\Leftrightarrow \cos 92+\cos 10=-\cos 88-\cos 170$
which is true because $\cos 92=\cos (180-88)=-\cos 88$
and $\cos 10=\cos (180-170)=-\cos 170$

4. Originally Posted by red_dog
$a^2-b^2+ac=\sin ^25-\sin ^249+\sin 5\sin 87=$
$=(\sin 5-\sin 49)(\sin 5+\sin 49)+\sin 5\sin 87=$
$=-2\sin 22\cos 27\cdot 2\sin 27\cos 22+\sin 5\sin 87=$
$=-\sin 44\sin 54+\sin 5\sin 87=-\frac{1}{2}(\cos 10-\cos 98-\cos 82+\cos 92)=$
$=-\frac{1}{2}(2\cos 51\cos 41-2\cos 90\cos 5)=-\cos 41\cos 51$

Now the equality becomes
$\sin 73=\frac{\cos 41\cos 51}{4\sin 5\cos 51\cos 41+\sin 5-\sin 49+\sin 87}$

We have $4\sin 5\cos 51\cos 41=2(\sin 56-\sin 46)\cos 41=2\sin 56\cos 41-2\sin 46\cos 41=$
$=\sin 15+\sin 97-\sin 5-\sin 87$
The equality becomes
$\sin 73=\frac{\cos 41\cos 51}{\sin 15+\sin 97-\sin 49}\Leftrightarrow$
$\Leftrightarrow \cos 41\cos 51=\sin 15\sin 73+\sin 97\sin 73-\sin 49\sin 73\Leftrightarrow$
$\Leftrightarrow \cos 92+\cos 10=\cos 58-\cos 88+\cos 24-\cos 170-\cos 24+\cos 122\Leftrightarrow$
$\Leftrightarrow \cos 92+\cos 10=-\cos 88-\cos 170$
which is true because $\cos 92=\cos (180-88)=-\cos 88$
and $\cos 10=\cos (180-170)=-\cos 170$
Dang!

-Dan