 # a+b+c=180° prove that cosa + cosb + cosc Tahmid Knox 2021-01-30 Answered

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

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

$a=60-{a}^{\prime }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}b$

$=60+{a}^{\prime }f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}0\le {a}^{\prime }<60º,\mathrm{cos}\left(0\right)=1,\mathrm{cos}\left(60\right)=0.5.$
Trig identity:

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

$=\mathrm{cos}A\mathrm{cos}+\mathrm{sin}A\mathrm{sin}B+\mathrm{cos}A\mathrm{cos}B-\mathrm{sin}A\mathrm{sin}B=2\mathrm{cos}A\mathrm{cos}B.$
We have:

$\mathrm{cos}\left(60-{a}^{\prime }\right)+\mathrm{cos}\left(60+{a}^{\prime }\right)-\mathrm{cos}\left(120\right)$

$=2\mathrm{cos}\left(60\right)\mathrm{cos}\left({a}^{\prime }\right)+0.5=\mathrm{cos}\left({a}^{\prime }\right)+0.5.\le 1.5.$
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 $\mathrm{cos}a+\mathrm{cos}b+1.$
Let $a=90-{a}^{\prime }$ and $b=90+{a}^{\prime },$

$0\mathrm{cos}\left(90-{a}^{\prime }\right)+\mathrm{cos}\left(90+{a}^{\prime }\right)+1=2\mathrm{cos}90\mathrm{cos}\left({a}^{\prime }\right)+1=1$

less than 1.5.