If x=cosA+ i sinA, y= cosB+ i sinB, z= cosG + i sinG and x+y+z=xyz, then prove that cos(B-G)+cos(G-A)+cos(A-B)=1

Question

Saurabh Koranglekar · Answer

Vikas TU · Answer

Dear student

by De Moivre's Theorem,
(cosa+cosb+cosc)+i(sina+sinb+sinc)=cos(a+b+c)+isin(a+b+c)
Equating real and imaginary parts,
cosa+cosb+cosc=cos(a+b+c)
and similarly for sine. Now,

(a−b)+(b−c)+(c−a)=0

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If x=cosA+ i sinA, y= cosB+ i sinB, z= cosG + i sinG and x+y+z=xyz, then prove that cos(B-G)+cos(G-A)+cos(A-B)=1

If x=cosA+ i sinA, y= cosB+ i sinB, z= cosG + i sinG and x+y+z=xyz, then prove that cos(B-G)+cos(G-A)+cos(A-B)=1

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If x=cosA+ i sinA, y= cosB+ i sinB, z= cosG + i sinG and x+y+z=xyz, then prove that cos(B-G)+cos(G-A)+cos(A-B)=1

If x=cosA+ i sinA, y= cosB+ i sinB, z= cosG + i sinG and x+y+z=xyz, then prove that cos(B-G)+cos(G-A)+cos(A-B)=1

2 Answers

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