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December 19th, 2015, 08:15 AM   #1
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From: Eski┼čehir

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This can also be written.. How?


I couldnt understand how this can be written? Which equation is used ?
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December 19th, 2015, 09:55 AM   #2
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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 10:10 AM.
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December 19th, 2015, 10:13 AM   #3
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Quote:
Originally Posted by Country Boy View Post
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?
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December 19th, 2015, 10:55 AM   #4
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$\begin{align*}2\cos^2B - 2\cos^2C &= \cos2B - \cos2C \\
&= -2\sin(B + C)\sin(B - C) \\
&= -2\sin(A)\sin(B - C) \ \ \text{(because } B + C = 180^\circ - A). \\
\end{align*}$

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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