Question

a+b+c=180° prove that cosa + cosb + cosc

Trigonometric Functions
ANSWERED
asked 2021-01-30

\(a+b+c=180^{\circ}\) prove that \(\displaystyle{\cos{{a}}}+{\cos{{b}}}+{\cos{{c}}}\)

Answers (1)

2021-01-31

If \(\displaystyle{a}={b}={c}={60}º,{\cos{{a}}}+{\cos{{b}}}+{\cos{{c}}}={3}\cdot\frac{{1}}{{2}}={1.5}.\)
\(\displaystyle{\cos{{c}}}={\cos{{\left({180}-{\left({a}+{b}\right)}\right)}}}=-{\cos{{\left({a}+{b}\right)}}}.\)
Let

\(\displaystyle{a}={60}-{a}'{\quad\text{and}\quad}{b}\)

\(={60}+{a}'{f}{\quad\text{or}\quad}{0}≤{a}'{<}{60}º,{\cos{{\left({0}\right)}}}={1},{\cos{{\left({60}\right)}}}={0.5}.\)
Trig identity:

\(\displaystyle{\cos{{\left({A}-{B}\right)}}}+{\cos{{\left({A}+{B}\right)}}}\)

\(={\cos{{A}}}{\cos{+}}{\sin{{A}}}{\sin{{B}}}+{\cos{{A}}}{\cos{{B}}}-{\sin{{A}}}{\sin{{B}}}={2}{\cos{{A}}}{\cos{{B}}}.\)
We have:

\(\displaystyle{\cos{{\left({60}-{a}'\right)}}}+{\cos{{\left({60}+{a}'\right)}}}-{\cos{{\left({120}\right)}}}\)

\(={2}{\cos{{\left({60}\right)}}}{\cos{{\left({a}'\right)}}}+{0.5}={\cos{{\left({a}'\right)}}}+{0.5}.≤{1.5}.\)
Also, as \(a+b\) approaches 0, the expression on the left \(< 1+1-1 <1\). 1 is less than 1.5.
As \(a+b\) approaches 180, the expression becomes \(\displaystyle{\cos{{a}}}+{\cos{{b}}}+{1}.\)
Let \(a=90-a'\) and \(b=90+a',\)

\(0\displaystyle{\cos{{\left({90}-{a}'\right)}}}+{\cos{{\left({90}+{a}'\right)}}}+{1}={2}{\cos{{90}}}{\cos{{\left({a}'\right)}}}+{1}={1}\)

less than 1.5.

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