The question is to prove: −a2+b2+2abcos⁡(α−β)&lt;acos⁡(α+θ)+bcos⁡(β+θ)&lt;a2+b2+2abcos⁡(α−β) I tried opening the brackets but Im

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The question is to prove:  −a2+b2+2abcos⁡(α−β)&amp;lt;acos⁡(α+θ)+bcos⁡(β+θ)&amp;lt;a2+b2+2abcos⁡(α−β)  I tried opening the brackets but Im

Foreckije · Accepted Answer

Note that  S=Re(aeα+θ+beβ+θ)  You can imagine these two complex numbers as vectors in the Argand plane.The resultant of these two vectors, can be found from the rule of addition of vectors and would have magnitude equal to a2+b2+2abcos⁡(α−β). Now, the real part of this resultant vector would be S. This would be its cosine component, which lies between −1 and 1. Hence the result holds up.

lalilulelo2k3eq · Accepted Answer

This is equivalent to proving  (acos⁡(α+θ)+bcos⁡(β+θ))2≤a2+b2+2abcos⁡(α−β)  ⇔a2(1−cos2(α+θ))+b2(1−cos2(β+θ))+2ab[cos⁡(α−β)−cos⁡(θ+α)cos⁡(θ+β)]≥0    [¬e: cos⁡(α−β)=cos⁡((θ+α)−(θ+β))=cos⁡(θ+α)(θ+β)+sin⁡(θ+α)(θ+β)]    ⇔a2sin2(α+θ)+b2sin2(β+θ)+2absin⁡(α+θ)sin⁡(β+θ)≥0    ⇔(asin⁡(α+θ)+bsin⁡(β+θ))2≥0, which is true    So, the given inequality holds

Vasquez · Accepted Answer

Imagine two rotating vectors of lengths a,b and initial directions α,β (with constant difference α−β). By the cosine rule in the triangle formed by these three vectors, the length of the sum is a2+b2+2abcos⁡(α−β) When the vectors rotate, the projection of the sum on the horizontal axis oscillates between plus and minus this value.

The question is to prove: -\sqrt{a^2+b^2+2ab \cos(\alpha-\beta)} < a \cos(\alpha+\theta)+b \cos

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