Question

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

Trigonometric Functions

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

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.