IIT JEE , AIEEE Forum - Mathematics , Trignometry

elastiboysai

Hii

I think u meant sin(A+B+C)???

cos A + cos (A+B) + cos (A+B+C) =0

and sin A + sin (A+B) + sin(A+B+C) = 0

Multiply second equation with i and add

u get

e^iA[1+e^iB+e^i(C+B)]=0

[1+e^iB+e^i(C+B)]=0------>[1]

call e^iB=z1 e^iB+C=z2

abv is poss only if z1=z2'(conj) [this follows bcoz of ur condn - B and C<pi]

and also 2Re(z1)=-1

so z1 shud b of the form -1/2+iy and z2= -1/2-iy

plug in and solve

u get B=C=2pi/3

Another simple way to see this problem is

1+W+W^2=0 ---->[2]

eqn [1] will fit in[2] perfect if e^iB=W,e^i(C+B)=W^2

hence B=2pi/3

B+C=4pi/3

but u can c clearly the second method is is just 4 understandin n not rigorous