Calculate cos 68 degrees + cos 112 degrees?

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You need to convert the sum of cosines into a product, such that:

`cos 68^o + cos 112^o = 2 cos((68^o + 112^o)/2)*cos((68^o - 112^o)/2)`

`cos 68^o + cos 112^o = 2 cos(180^o/2)cos(-44^o/2)`

`cos 68^o + cos 112^o = 2 cos(90^o)*cos(-22^o)`

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

`2 cos(90^o)*cos(-22^o) = 0`

Hence, evaluating the summation of cosines, yields `cos 68^o + cos 112^o = 0.`

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