why is cos θ =e^iθ+e^-iθ/2
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b.tech 2nd year!!!
so what is the answer anyway???
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eix = cosx + isinx (1)
Now substituting x=-x, we get
e-ix = cos(-x) + isin(-x)
e-ix = cosx - isinx (2)
Now adding (1) and (2)
eix + e-ix= cosx + isinx + cosx - isinx
eix + e-ix= 2*cosx (3)
Now dividing (3) by 2 , we get
( eix + e-ix ) / 2= cosx
Now substituting x=-x, we get
e-ix = cos(-x) + isin(-x)
e-ix = cosx - isinx (2)
Now adding (1) and (2)
eix + e-ix= cosx + isinx + cosx - isinx
eix + e-ix= 2*cosx (3)
Now dividing (3) by 2 , we get
( eix + e-ix ) / 2= cosx
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