Find the values of α and β, 0<α, β < Π/2, satisfying the eqn -
cos α cos β cos (α+β) = -1/8.
1HINT:
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 ......
12 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
106again u cant equate such terms.. what is the proof that cos(2@) has to be -1/2 ?
1Well .... My solution satisfies the expr. Why don't u come up with a better ans.??????!!!!!!
106But 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
10<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.
1I never said that these r the only values. This might be one possible solution to the problem.
62precisely Asish's point..
I dont see these getting solved :(
Unless one is given in terms of the other!
1i got the point!..
make quadratic in cos(α+β)... then
apply the condition on its discriminant to be greater than or equal to zero.!!