1+cos =? And 1-cos =?

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You also may solve this problem using the following approach, hence, first, replace `cos 0` for the value `1` and then, convert the summation or difference of cosines into a product, such that:

`1 + cos theta = cos 0 + cos theta = 2 cos((0 + theta)/2)*cos((0 - theta)/2)`

`1 + cos theta = 2 cos((theta)/2)*cos((-theta)/2)`

Using the property `cos (-theta) = cos theta` , yields:

`1 + cos theta = 2 cos((theta)/2)*cos((theta)/2)`

`1 + cos theta = 2 cos^2((theta)/2)`

Evaluating the difference `1 - cos theta` , yields:

`1 - cos theta = 2 sin((0 + theta)/2)*sin((theta - 0)/2)`

`1 - cos theta = 2 sin^2((theta)/2)`

Hence, evaluating the given summation and difference, yields `1 + cos theta = 2 cos^2((theta)/2)` and `1 - cos theta = 2 sin^2((theta)/2).`

1 + cos theta =

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Using half angle trigonometric identities:

`tan(A/2)=sinA/(1+cosA)=(1-cosA)/sinA`

Therefore:

`1+cosA=sinA/tan(A/2)`

`1-cosA=sinAtan(A/2)`

Verification:

If you multiply `(1+cosA)` by `(1-cosA)`, the result is:

`1-cos^2A=(sinA)(sinA)(tan(A/2))/tan(A/2)`

` ` Thus: `1-cos^2A=sin^2A`

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