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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: 896 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
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Joined: Oct 2015
From: Eskişehir

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Quote:
 Originally Posted by Country Boy You need to use some trig identities: $\displaystyle \sin(A+ B)= \sin(A)\cos(B)+ \cos(A)\sin(B)$ $\displaystyle \cos(A+ B)= \cos(A)\cos(B)- \sin(A)\sin(B)$ $\displaystyle \sin(2A)= 2\sin(A)\cos(A)$ $\displaystyle \cos(2A)= \cos^2(A)- \sin^2(A)$
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: 20,274 Thanks: 1959 \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. Thanks from yuceelly

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