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

Carla Murphy 2022-01-02 Answered
The question is to prove:
\(\displaystyle-\sqrt{{{a}^{{2}}+{b}^{{2}}+{2}{a}{b}{\cos{{\left(\alpha-\beta\right)}}}}}{ < }{a}{\cos{{\left(\alpha+\theta\right)}}}+{b}{\cos{{\left(\beta+\theta\right)}}}{ < }\sqrt{{{a}^{{2}}+{b}^{{2}}+{2}{a}{b}{\cos{{\left(\alpha-\beta\right)}}}}}\)
I tried opening the brackets but I'm not succeeding. I even tried to differentiate the function. But I'm not getting any closer to the answer.

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Expert Answer

Foreckije
Answered 2022-01-03 Author has 953 answers
Note that
\(\displaystyle{S}={R}{e}{\left({a}{e}^{{\alpha+\theta}}+{b}{e}^{{\beta+\theta}}\right)}\)
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 \(\displaystyle\sqrt{{{a}^{{2}}+{b}^{{2}}+{2}{a}{b}{\cos{{\left(\alpha-\beta\right)}}}}}\). 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.
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lalilulelo2k3eq
Answered 2022-01-04 Author has 2680 answers
This is equivalent to proving
\(\displaystyle{\left({a}{\cos{{\left(\alpha+\theta\right)}}}+{b}{\cos{{\left(\beta+\theta\right)}}}\right)}^{{2}}\leq{a}^{{2}}+{b}^{{2}}+{2}{a}{b}{\cos{{\left(\alpha-\beta\right)}}}\)
\(\displaystyle\Leftrightarrow{a}^{{2}}{\left({1}-{{\cos}^{{2}}{\left(\alpha+\theta\right)}}\right)}+{b}^{{2}}{\left({1}-{{\cos}^{{2}}{\left(\beta+\theta\right)}}\right)}+{2}{a}{b}{\left[{\cos{{\left(\alpha-\beta\right)}}}-{\cos{{\left(\theta+\alpha\right)}}}{\cos{{\left(\theta+\beta\right)}}}\right]}\geq{0}\)
\(\displaystyle{\left[\neg{e}:\ {\cos{{\left(\alpha-\beta\right)}}}={\cos{{\left({\left(\theta+\alpha\right)}-{\left(\theta+\beta\right)}\right)}}}={\cos{{\left(\theta+\alpha\right)}}}{\left(\theta+\beta\right)}+{\sin{{\left(\theta+\alpha\right)}}}{\left(\theta+\beta\right)}\right]}\)
\(\displaystyle\Leftrightarrow{a}^{{2}}{{\sin}^{{2}}{\left(\alpha+\theta\right)}}+{b}^{{2}}{{\sin}^{{2}}{\left(\beta+\theta\right)}}+{2}{a}{b}{\sin{{\left(\alpha+\theta\right)}}}{\sin{{\left(\beta+\theta\right)}}}\geq{0}\)
\(\displaystyle\Leftrightarrow{\left({a}{\sin{{\left(\alpha+\theta\right)}}}+{b}{\sin{{\left(\beta+\theta\right)}}}\right)}^{{2}}\geq{0}\), which is true
So, the given inequality holds
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Vasquez
Answered 2022-01-08 Author has 9499 answers

Imagine two rotating vectors of lengths a,b and initial directions \(\alpha,\beta\) (with constant difference \(\alpha-\beta\)). By the cosine rule in the triangle formed by these three vectors, the length of the sum is
\(\sqrt{a^2+b^2+2ab \cos(\alpha-\beta)}\)
When the vectors rotate, the projection of the sum on the horizontal axis oscillates between plus and minus this value.

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