Math Tutor Online –  In any triangle ABC, prove that

1. A cosA + b cosB + c cosC = 4RsinA sinB sinC
2. sinA + sinB + sinC = S/R

 

Let’s see the solution of the following

1. lets first have left hand side equation which is

 

A cosA + b cosB + c cosC
= 2R sin A cosA + 2R sin B cos B + 2R sinC cos C

This is because as we know a = 2R sinA, b=2RsinB, c = 2RsinC

So now taking R common and converting into formula that is 2 sin A cosA = sin2A

= R (sin2A + sin2B + sin2C)

Again by applying formula

= R (2sin (A+B) cos(A-B) + 2 sinC cos C )

= R (2sin (π –C)cos(A-B)+ 2 sinC cos C )

= R (2sinC cos(A-B)+ 2 sinC cos C )

Now taking 2 sin C as common we have

= 2RsinC (cos(A-B)+ cos C )

= 2RsinC (cos(A-B)+ cos(π – (A+B ))

= 2RsinC (cos(A-B)- cos (A+B ))

= 2RsinC (2sinA sin B)

= 4R sin A sin B sin C = right hand side

Hence proved

1. sinA + sinB + sinC = S/R

here we should know things like

R = a/2sinA= b/2sinB = c/2sinC

 

sinA = a/2R, sinB = b/2R, sinC= c/2R

Just by putting these values in left hand side we get

a/2R +b/2R+c/2R =  a+b+c/2R = 2S/2R = S/R = right hand side

Hence proved.