213
Question
Find sinθ+ sin(π+θ)+ sin(2π+θ)+ sin(3π+θ)+⋯ upto 2021 terms.
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Solution
sinθ+sin(π+θ)+sin(2π+θ)+......+sin(2021π+θ)Nowsin(π+θ)=sin(3π+θ)=.........=sin[(2n+1)π+θ]=−sinθsin(2π+θ)=sin(4π+θ)=.........=sin(2nπ+θ)=sinθ∴sin(2021π+θ)+sin(2020π+θ)=0sin(2019π+θ)+sin(2018π+θ)=0.....sin(3π+θ)+sin(2π+θ)=0sin(π+θ)+sinθ=0∴sum=0
Now
sin(π+θ)=sin(3π+θ)=.........=sin[(2n+1)π+θ]=−sinθsin(2π+θ)=sin(4π+θ)=.........=sin(2nπ+θ)=sinθ∴sin(2021π+θ)+sin(2020π+θ)=0sin(2019π+θ)+sin(2018π+θ)=0.....sin(3π+θ)+sin(2π+θ)=0sin(π+θ)+sinθ=0∴sum=0
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