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December 19th, 2015, 09:15 AM  #1 
Newbie Joined: Oct 2015 From: Eskişehir Posts: 11 Thanks: 0  This can also be written.. How? I couldnt understand how this can be written? Which equation is used ? 
December 19th, 2015, 10:55 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
You need to use some trig identities: $\displaystyle \sin(A+ B)= \sin(A)\cos(B)+ \cos(A)\sin(B)$ $\displaystyle \cos(A+ B)= \cos(A)\cos(B) \sin(A)\sin(B)$ $\displaystyle \sin(2A)= 2\sin(A)\cos(A)$ $\displaystyle \cos(2A)= \cos^2(A) \sin^2(A)$ Last edited by skipjack; December 19th, 2015 at 11:10 AM. 
December 19th, 2015, 11:13 AM  #3 
Newbie Joined: Oct 2015 From: Eskişehir Posts: 11 Thanks: 0  Yeah I know but I couldnt use actually Can you write how to do it?

December 19th, 2015, 11:55 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,865 Thanks: 1833 
$\begin{align*}2\cos^2B  2\cos^2C &= \cos2B  \cos2C \\ &= 2\sin(B + C)\sin(B  C) \\ &= 2\sin(A)\sin(B  C) \ \ \text{(because } B + C = 180^\circ  A). \\ \end{align*}$ Hence $2\cos{A}(\cos^2B  \cos^2C) = 2\sin{A}\cos{A}\sin(B  C) = \sin2A\sin(B  C)$, etc. 

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