Real part of sin(a+i B) is???
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Step-by-step explanation:
sin(α+iβ)=sinα.cos(iβ)+cosα.sin(iβ)
cos(iβ)=1−
2!
(iβ)
2
+
4!
(iβ)
4
−...
=1+
2!
β
2
+
4!
β
4
+...
2
e
β
+e
−β
=
2
1
[(1+
1!
β
+
2!
β
2
+...)+(1−
1!
β
+
2!
β
2
−...)]
=1+
2!
β
2
+
4!
β
4
+...=cos(iβ)
sin(iβ)=iβ−
3!
(iβ)
3
+
5!
(iβ)
5
−...
=i[β+
3!
β
3
+
5!
β
5
−...]=
2
i[e
β
+e
−β
]
∴sin(α+iβ)=sinα.
2
e
β
+e
−β
+cosα.i(
2
e
β
+e
−β
)
∴ Real part of sin(α+iβ)=sinα.
2
e
β
+e
−β
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