Carla Murphy

2022-01-02

The question is to prove:

$-\sqrt{{a}^{2}+{b}^{2}+2ab\mathrm{cos}(\alpha -\beta )}<a\mathrm{cos}(\alpha +\theta )+b\mathrm{cos}(\beta +\theta )<\sqrt{{a}^{2}+{b}^{2}+2ab\mathrm{cos}(\alpha -\beta )}$

I tried opening the brackets but Im

I tried opening the brackets but Im

Foreckije

Beginner2022-01-03Added 32 answers

Note that

$S=Re(a{e}^{\alpha +\theta}+b{e}^{\beta +\theta})$

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$\sqrt{{a}^{2}+{b}^{2}+2ab\mathrm{cos}(\alpha -\beta )}$ . 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.

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

lalilulelo2k3eq

Beginner2022-01-04Added 38 answers

This is equivalent to proving

${(a\mathrm{cos}(\alpha +\theta )+b\mathrm{cos}(\beta +\theta ))}^{2}\le {a}^{2}+{b}^{2}+2ab\mathrm{cos}(\alpha -\beta )$

$\iff {a}^{2}(1-{\mathrm{cos}}^{2}(\alpha +\theta ))+{b}^{2}(1-{\mathrm{cos}}^{2}(\beta +\theta ))+2ab[\mathrm{cos}(\alpha -\beta )-\mathrm{cos}(\theta +\alpha )\mathrm{cos}(\theta +\beta )]\ge 0$

$[\mathrm{\neg}e:\text{}\mathrm{cos}(\alpha -\beta )=\mathrm{cos}((\theta +\alpha )-(\theta +\beta ))=\mathrm{cos}(\theta +\alpha )(\theta +\beta )+\mathrm{sin}(\theta +\alpha )(\theta +\beta )]$

$\iff {a}^{2}{\mathrm{sin}}^{2}(\alpha +\theta )+{b}^{2}{\mathrm{sin}}^{2}(\beta +\theta )+2ab\mathrm{sin}(\alpha +\theta )\mathrm{sin}(\beta +\theta )\ge 0$

$\iff {(a\mathrm{sin}(\alpha +\theta )+b\mathrm{sin}(\beta +\theta ))}^{2}\ge 0$ , which is true

So, the given inequality holds

So, the given inequality holds

Vasquez

Expert2022-01-08Added 669 answers

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