Question

please give a correct solution for this question.​

Answer 1

Step-by-step explanation:

sinθ+sin(π+θ)+sin (2π+θ)+......+sin(2021π+θ)

Now

sin(π+θ)=sin(3π+θ)=.........=sin[(2n+1)π+θ]=−sinθ

sin(2π+θ)=sin(4π+θ)=.........=sin(2nπ+θ)=sinθ

∴sin(2021π+θ)+sin(2020π+θ)=0

sin(2019π+θ)+sin(2018π+θ)=0

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.sin(3π+θ)+sin(2π+θ)=0

sin(π+θ)+sinθ=0

∴sum=0

Hope it helps ...

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