$\cos{40^\circ}$ $+$ $\cos{80^\circ}$ $+$ $\cos{160^\circ}$

The values of cosine of these three angles are unknown. So, it cannot solved directly by substituting values of them in this trigonometric expression. So, an alternative approach should be used to simplify it.

All three terms are cosine functions. So, there is no problem to apply sum to product transformation identities.

$\cos{\alpha}+\cos{\beta}$ $=$ $2\cos{\Bigg(\dfrac{\alpha+\beta}{2}\Bigg)}\cos{\Bigg(\dfrac{\alpha-\beta}{2}\Bigg)}$

In this math problem, $\alpha = 40^\circ$ and $\beta = 80^\circ$. Substitute them and simplify the trigonometric expression.

$=$ $2\cos{\Bigg(\dfrac{40^\circ + 80^\circ}{2}\Bigg)}\cos{\Bigg(\dfrac{40^\circ – 80^\circ}{2}\Bigg)}$ $+$ $\cos{160^\circ}$

$=$ $2\cos{\Bigg(\dfrac{120^\circ}{2}\Bigg)}\cos{\Bigg(\dfrac{-40^\circ}{2}\Bigg)}$ $+$ $\cos{160^\circ}$

$=$ $2\cos{60^\circ}\cos{(-20^\circ)}$ $+$ $\cos{160^\circ}$

As per cos of negative angle identity, cosine of a negative angle is equal to the cosine of same positive angle.

$=$ $2\cos{60^\circ}\cos{20^\circ}$ $+$ $\cos{160^\circ}$

$=$ $2 \times \dfrac{1}{2} \times \cos{20^\circ}$ $+$ $\cos{160^\circ}$

The value of cosine of $60^\circ$ is equal to $\dfrac{1}{2}$. Replace $\cos{60^\circ}$ by this value to go ahead in simplifying this trigonometry problem.

$=$ $\dfrac{2}{2} \times \cos{20^\circ}$ $+$ $\cos{160^\circ}$

$=$ $\require{cancel} \dfrac{\cancel{2}}{\cancel{2}} \times \cos{20^\circ}$ $+$ $\cos{160^\circ}$

$=$ $\cos{20^\circ}$ $+$ $\cos{160^\circ}$

Sum to product transforming trigonometric identity can be used one more time but it can also be solved mathematically by trigonometric ratios of allied angles.

$=$ $\cos{20^\circ}$ $+$ $\cos{(180^\circ -20^\circ)}$

$=$ $\cos{20^\circ}$ $-\cos{20^\circ}$

$= 0$

$\,\,\, \therefore \,\,\,\,\,\,$ $\cos{40^\circ}$ $+$ $\cos{80^\circ}$ $+$ $\cos{160^\circ} = 0$

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