# Calculate `cos 72^o + cos108^o` ?

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You may convert the summation of cosines in a product, using the following formula, such that:

`cos a + cos b = 2 cos(a/2 + b/2)cos(a/2 - b/2)`

Reasoning by analogy, yields:

`cos 72^o + cos 108^o = 2cos((72^o)/2 + (108^o)/2)cos((72^o)/2 - (108^o)/2)`

`cos 72^o + cos 108^o = 2cos(36^o + 54^o)cos(36^o - 54^o)`

`cos 72^o + cos 108^o = 2cos(90^o)cos(-18^o)`

Since `cos(90^o) = 0` and using zero product rule yields:

`cos 72^o + cos 108^o = 0`

**Hence, evaluating the summation of cosines, converting it into a product, yields **`cos 72^o + cos 108^o = 0.`

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