help!

HINT:

Multiply the expr. on both sides by 2

2 cos α cos β cos (α+β) = -14

={ cos(α+β) + cos(α-β) } cos(α+β) = -14

Now proceed ......

what to do after dis???

2 cos α cos β cos (α+β) = -14 ....... (1)

={ cos (α+β) + cos (α-β) } cos (α+β) = -14 ......(1) * 2

=2 cos² (α+β) + 2 cos (α+β) cos (α-β) = -12 .... (1)

=2 cos² (α+β) + cos 2α + cos 2β = -12 .... (1)

subtracting -1 on both sides of (1) .....

=cos 2(α+β) + cos 2α + cos 2β = -32 .... (1)

=cos 2(α+β) + cos 2α + cos 2β = -12 - 12 - 12 .... (1)

cos 2α = -1/2
cos 2α = -cos 60
= cos 120
2α = 120
α = 60
Similarly ... β=60

again u cant equate such terms.. what is the proof that cos(2@) has to be -1/2 ?

Well .... My solution satisfies the expr. Why don't u come up with a better ans.??????!!!!!!

But u cannot be sure (from ur steps) that these are the only two values of α,β that satisfy. ..

I hve not yet been able to get an answer to this question yet . I am just pointing out why your solution is not valid

0<α, β<Π/2

0<2α, 2β<Î

cos 2α, cos 2β lies between -1 and 1

Well I can't think of any easier way of breakup of -32
than what I've done earlier.

I never said that these r the only values. This might be one possible solution to the problem.