Answer
Let S be the sum of series
S=sinα−sin(α+β)+sin(α+2β)−sin(α+3β)+....
Multiplying with 2sin(
2
β
) on both sides
2sin(
)S=2sin(
)sinα−2sin(
)sin(α+β)+2sin(
)sin(α+2β)−....
)S=cos(α+
)−cos(α−
)+cos(α+
3β
)−cos(α+
5β
)
−cos(α+
)+...+cos(α+(n−1)β+
)−cos(α+(n−1)β−
)S=cos(α+(n−1)β+
)S=−2sin(α+
(n−1)β
)sin(
nβ
S=−
sin(
sin(α+
Step-by-step explanation:
Answer:
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Answers & Comments
Answer
Let S be the sum of series
S=sinα−sin(α+β)+sin(α+2β)−sin(α+3β)+....
Multiplying with 2sin(
2
β
) on both sides
2sin(
2
β
)S=2sin(
2
β
)sinα−2sin(
2
β
)sin(α+β)+2sin(
2
β
)sin(α+2β)−....
2sin(
2
β
)S=cos(α+
2
β
)−cos(α−
2
β
)+cos(α+
2
3β
)−cos(α+
2
β
)+cos(α+
2
5β
)
−cos(α+
2
3β
)+...+cos(α+(n−1)β+
2
β
)−cos(α+(n−1)β−
2
β
)
2sin(
2
β
)S=cos(α+(n−1)β+
2
β
)−cos(α−
2
β
)
2sin(
2
β
)S=−2sin(α+
2
(n−1)β
)sin(
2
nβ
)
S=−
sin(
2
β
)
sin(α+
2
(n−1)β
)sin(
2
nβ
)
Step-by-step explanation:
Answer:
hello r u malayali I don't know why u go llow me without any reason that's why asking