α and β do not differ by an even multiple of π. If θ satisfies cosαcosθ+sinαsinθ=cosβcosθ+sinβsinθ=1 then show that cosαcosβcos2θ+sinαsinβsin2θ+1=0
I wish to solve this problem using some elegant method, preferably complex numbers.
I've tried using the fact that α and β satisfy an equation of the form cosx/cosθ+sinx/sinθ=1, and got the required result. See my solution here: https://www.pdf-archive.com/2017/07/01/solution
I'm guessing there's an easier way to go about it. Thanks in advance!
Answer
Well, I'm not sure I can do that in a very elegant way, but it might be shorter. I'm using addition theorems, and the identity sinx−siny=2sinx−y2cosx+y2 following immediately from them.
Multiplying the given equations by sinθcosθ, we get
sin(α+θ)=sinθcosθ=sin(β+θ),
but 0=sin(α+θ)−sin(β+θ)=2sinα−β2cos(α+β2+θ). The first factor is ≠0 by assumption, so cos(α+β2+θ)=0. Multiplying by 2sin(α+β2−θ) and using the above identity, you get sin(α+β)−sin2θ=0, and this means (using sin2θ=2sinθcosθ and dividing by sinθcosθ)
sinαcosβsinθcosθ+sinβcosαsinθcosθ=2. Now you have
(cosαcosθ+sinαsinθ)(cosβcosθ+sinβsinθ)−(sinαcosβsinθcosθ+sinβcosαsinθcosθ)=1⋅1−2=−1, and that gives your required result after simplifying.
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