Get the solution to "a+b+c=180∘ prove that cos⁡a+cos⁡b+cos⁡c"

High School
Geometry
Trigonometry
Trigonometric Functions

Tahmid Knox

Answered question

2021-01-30

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

Answer & Explanation

Gennenzip

Skilled2021-01-31Added 96 answers

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

$a = 60 - a^{'} and b$

$= 60 + a^{'} f or 0 \leq a^{'} < 60 º, \cos (0) = 1, \cos (60) = 0.5 .$
Trig identity:

$\cos (A - B) + \cos (A + B)$

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

$\cos (60 - a^{'}) + \cos (60 + a^{'}) - \cos (120)$

$= 2 \cos (60) \cos (a^{'}) + 0.5 = \cos (a^{'}) + 0.5 . \leq 1.5 .$
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 $\cos a + \cos b + 1 .$
Let $a = 90 - a^{'}$ and $b = 90 + a^{'},$

$0 \cos (90 - a^{'}) + \cos (90 + a^{'}) + 1 = 2 \cos 90 \cos (a^{'}) + 1 = 1$

less than 1.5.

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