1. ## Trigonometry Identity

I need help for my homework, thanks.

1) $\tan(X+Y) + \tan(X-Y) = \frac{2\sin(2X)}{\cos(2X) + \cos(2Y)}$

2) If A + B + C = 180 degrees, prove that $\cos(A) + \cos(B) + \cos(C) = 1 + 4\sin{\frac{A}{2}}\sin{\frac{b}{2}}\sin{\frac{c}{2 }}$

Thank you.

2. Hello, elitewarr!

For the first one, we need three identities:

. . $\tan\theta \:=\:\frac{\sin\theta}{\cos\theta}$

. . $\sin A\cos B - \sin B\cos A \:=\:\sin(A - B)$

. . $\cos A\cos B \:=\:\tfrac{1}{2}\bigg[\cos(A+B) + \cos(A-B)\bigg]$

$1)\;\;\tan(x+y) + \tan(x-y) \:= \:\frac{2\sin2x}{\cos2x + \cos2y}$

$\tan(x+y) + \tan(x-y) \:=\:\frac{\sin(x+y)}{\cos(x+y)} + \frac{\sin(x-y)}{\cos(x-y)}$

. . . . . $= \;\frac{\sin(x+y)\cos(x-y) + \sin(x-y)\cos(x+y)}{\cos(x+y)\cos(x-y)}$

. . . . . $= \;\frac{\sin\bigg[(x+y) + (x-y)\bigg]}{\frac{1}{2}\bigg[\cos[(x+y)+ (x-y)] + \cos[(x+y)-(x-y)]\bigg]}$

. . . . . $= \;\frac{\sin2x}{\frac{1}{2}(\cos2x + \cos 2y)}$

. . . . . $= \;\frac{2\sin2x}{\cos2x+\cos2y}$

3. OHH!! Never did I ever thought of using sin(a-b).. Haha... Thanks