1this problem should be solved using Viète's Laws...........
let t1=cot2 α
t2=cot2 β
t3 =cot2 γ
2+t1+t2+t3=t1t2t3
let there be a cubic whose roots are t1,t2,t3
ax3+bx2+cx+d=0
-ba+2=-da
-d=2a-b
ax3+bx2+cx+b-2a=0
now
cos 2α=t1-1t1+1
now instead of x in cubic replace it with 1+x1-x
new cubic : a(1+x1-x)3+b(1+x1-x)2+c(1+x1-x)+b-2a=0
now the new cubic will have roots as
cos 2α,cos 2β,cos 2γ
and now no need to evaluate whole cubic just find coefficient of x3 and x2
now
Σcos 2α =-coefficient of x2coefficient of x3=3a-2b+c3a-2b+c=1
39Don't think of a pink on your post as some coveted and treasured possession. It means nothing. It is only used to highlight posts which are informative or posts with nice answers. 500 pink posts cannot guarantee you IIT. And this is a pretty good solution, never thought we could apply Viete's rules here.
1
Sonne
·2010-07-13 23:02:39
hey pritish fyi my iit-jee is over ............
39
Pritish Chakraborty
·2010-07-13 23:13:23
And fyi I'm the one who pinked your post :)
1
Sonne
·2010-07-16 09:13:29
is there any other proof ?
341there's always brute force manipulation :D
Σ\cos 2 \alpha = \sum \frac{1-\tan^2 \alpha}{1+\tan^2 \alpha} = 3 - 2\sum \frac{\tan^2 \alpha}{1+\tan^2 \alpha}
3 - \frac{2\sum \tan^2 \alpha(1+\tan^2 \beta)(1+\tan^2 \gamma)}{(1+\tan^2 \alpha)(1+\tan^2 \beta)(1+\tan^2 \gamma)}
= 3 - \frac{2\left(\sum \tan^2 \alpha+2\sum \tan^2 \alpha \tan^2 \beta+3 \tan^2 \alpha \tan^2 \beta \tan^2 \gamma\right)}{(1+\tan^2 \alpha)(1+\tan^2 \beta)(1+\tan^2 \gamma)}
= 3 - \frac{2\left(1+\sum \tan^2 \alpha+\sum \tan^2 \alpha \tan^2 \beta+ \tan^2 \alpha \tan^2 \beta \tan^2 \gamma\right)}{(1+\tan^2 \alpha)(1+\tan^2 \beta)(1+\tan^2 \gamma)}
(Using the given relation)
3 - \frac{2(1+\tan^2 \alpha)(1+\tan^2 \beta)(1+\tan^2 \gamma))}{(1+\tan^2 \alpha)(1+\tan^2 \beta)(1+\tan^2 \gamma)}
= 3-2 = \boxed {1}
