trigonometry

[man111]

if [tex]sinA+sinB+sinC = cosA+cosB+cosC = 0[/tex] then show that

[tex]cos2A+cos2B+cos2C = sin2A+sin2B+sin2C = 0[/tex]

[martin123456]

[tex]\sin{A}+\sin{B}=-\sin{C}[/tex]
[tex]\cos{A}+\cos{B}=-\cos{C}[/tex]
We divide them and get [tex]tg\frac{A+B}{2}=tgC[/tex], so [tex]\frac{A+B}{2}-C=\pi k[/tex](1)
Multiplying the above identities, we get [tex]\sin{C}\cos{C}=\sin{A}\cos{A}+\sin{B}\cos{B}+\sin{A}\cos{B}+\sin{B}\cos{A}[/tex] and so [tex]\sin{2C}=\sin{2A}+\sin{2B}+2\sin{(A+B)[/tex]. We add [tex]\sin{2C}[/tex] and we get that [tex]4\sin{(C-\frac{A+B}{2})}\cos{(C+\frac{A+B}{2})}=\sum{\sin{2A}}[/tex]. From (1) the RHS is 0.

[man111]

Thanks martin.

How can we solve this by using Complex no.

[allier]

[tex]a=cosA+isinA[/tex]
[tex]b=cosB+isinB[/tex]
[tex]c=cosC+isinC[/tex]
a,b,c all lie on the unit circle and by the initial condition, a+b+c=0, so a,b,c form a triangle whose centroid coincides with the center of its circumscribed circle. Thus, a,b,c form an equilateral triangle. Now, it's easy to see that [tex]a^2,b^2,c^2[/tex] also form an equilateral triangle - for example, [tex]|a^2-b^2|=|a-b||a+b|=|a-b||-c|=|a-b|[/tex]. Thus, [tex]a^2+b^2+c^2=0[/tex], which implies [tex]cos2A+cos2B+cos2C=0[/tex] and [tex]sin2A+sin2B+sin2C=0[/tex].