Check out some similar questions!

saylow · Jul 26, 2010, 02:11 PM

verify that sin(A)+sin(B)+sin(C) = sin(A+B+C) + 4sin((B+C)/2)sin((A+C)/)sin((A+B)/2)

Unknown008 · Jul 27, 2010, 07:47 AM

Ok...

$sinA + sinB + sinC \equiv sin(A+B+C) + 4sin(\frac{A+B}{2})sin(\frac{B+C}{2})sin(\frac{A+C }{2})$

I think it's better if we start on the right part.

$sin(A+B+C) + 4sin(\frac{A+B}{2})sin(\frac{B+C}{2})sin(\frac{A+C }{2})$

Breaking sin(A+B+C);

$sin(A+B+C) = sin(A+B)cosC + cos(A+B)sinC$

$= (sinAcosB + cosAsinB)cosC + (cosAcosB - sinAsinB)sinC$

$= sinAcosBcosC + cosAsinBcosC + cosAcosBsinC - sinAsinBsinC$

Breaking the other part, using the identity $sinA = 2sin(\frac{A}{2})cos(\frac{A}{2})$

$4sin(\frac{A+B}{2})sin(\frac{B+C}{2})sin(\frac{A+C }{2}) = 4(\frac{sin(A+B)}{2cos(A+B)})(\frac{sin(B+C)}{2cos (B+C)})(\frac{sin(A+C)}{2cos(A+C)})$

$= \frac12 (\frac{sinAcosB+ cosAsinB}{cosAcosB - sinAsinB})(\frac{sinBcosC+cosBsinC}{cosBcosC-sinBsinC})(\frac{sinAcosC+cosAsinC}{cosAcosC-sinAsinC})$

Phew! I give up here. That's terrifically long. I expanded giving 8 dissimilar terms in the numerator and denominator (actually, there are two sinAcosAsinBcosBsinCcosC which becomes 2sinAcosAsinBcosBsinCcosC in the numerator) . Anyone wants to continue? :o

psychjr. · Jul 20, 2011, 01:49 PM

cos^2((Î¸)/(2))=(secÎ¸+1)/(2secÎ¸)

Ezemadu123 · Jul 30, 2011, 10:01 AM

Expand (1-e^i(2theta))/(1-e^i(theta)

edward.elric · Sep 29, 2011, 04:06 AM

sin^2A/cosn=secA-cosA