Question

2cos@+2sin@ is equals to??
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Answer 1

Step-by-step explanation:

2sin

2

β+4cos(α+β).sinαsinβ+2cos

2

(α+β)−1

=2sin

2

β+2cos(α+β)[2sinαsinβ+cos(α+β)]−1

=2sin

2

β+2cos(α+β)[2sinαsinβ+cosαcosβ−sinαsinβ]−1

=2sin

2

β+2cos(α+β)cos(α−β)−1

=2sin

2

β+cos2α+cos2β−1

=−cos2β+cos2α+cos2β

=cos2α