Question

4cosA/2cosB/2cosC/2=?

Answer 1

Answer:

sinA+sinB+sinC=4cos2Acos2Bcos2C

Step-by-step explanation:

In ΔABC , Show that

sinA+sinB+sinC=4\:cos\frac{A}{2}\:cos\frac{B}{2}\:cos\frac{C}{2}sinA+sinB+sinC=4cos2Acos2Bcos2C

\begin{gathered}Formula\:used:\\\\sinC+sinD=2\:sin(\frac{C+D}{2})\:cos(\frac{C-D}{2})\\\\sinA=2\:sin\frac{A}{2}\:cos\frac{A}{2}\end{gathered}Formulaused:sinC+sinD=2sin(2C+D)cos(2C−D)sinA=2sin2Acos2A

In ΔABC , A+B+C=\piA+B+C=π

\begin{gathered}sinA+sinB+sinC\\\\=2\:sin(\frac{A+B}{2})cos(\frac{A-B}{2})+2\:sin\frac{C}{2}\:cos\frac{C}{2}\\\\=2\:sin(\frac{\pi-C}{2})cos(\frac{A-B}{2})+2\:sin\frac{C}{2}\:cos\frac{C}{2}\end{gathered}sinA+sinB+sinC=2sin(2A+B)cos(2A−B)+2sin2Ccos2C=2sin(2π−C)cos(2A−B)+2sin2Ccos2C

\begin{gathered}=2\:sin(\frac{\pi}{2}-\frac{C}{2})cos(\frac{A-B}{2})+2\:sin\frac{C}{2}\:cos\frac{C}{2}\\\\=2\:cos\frac{C}{2}cos(\frac{A-B}{2})+2\:sin\frac{C}{2}\:cos\frac{C}{2}\\\\=2\:cos\frac{C}{2}[cos(\frac{A-B}{2})+\:sin\frac{C}{2}]\end{gathered}=2sin(2π−2C)cos(2A−B)+2sin2Ccos2C=2cos2Ccos(2A−B)+2sin2Ccos2C=2cos2C[cos(2A−B)+sin2C]

\begin{gathered}=2\:cos\frac{C}{2}[cos(\frac{A-B}{2})+\:sin\frac{\pi-(A+B)}{2}]\\\\=2\:cos\frac{C}{2}[cos(\frac{A-B}{2})+\:sin(\frac{\pi}{2}-\frac{A+B}{2})]\\\\=2\:cos\frac{C}{2}[cos(\frac{A-B}{2})+cos(\frac{A+B}{2})]\end{gathered}=2cos2C[cos(2A−B)+sin2π−(A+B)]=2cos2C[cos(2A−B)+sin(2π−2A+B)]=2cos2C[cos(2A−B)+cos(

Answer 2

Step-by-step explanation:

SinA + SinB + SinC = 4cosAcosBcosC

4cosA/2cosB/2cosC/2 = SinA/2 + SinB/2 + SinC/2