Evaluating \frac{1}{\sin(2x)} + \frac{1}{\sin(4x)} + \frac{1}{\sin(8x)} + \frac{1}{\sin(16x)}

segnverd3a

segnverd3a

Answered question

2022-01-30

Evaluating 1sin(2x)+1sin(4x)+1sin(8x)+1sin(16x)

Answer & Explanation

enveradapb

enveradapb

Beginner2022-01-31Added 13 answers

sin(AB)=sinAcosBcosAsinB
1sin2x=sin(2xx)sin2xsinx=cotxcot2x
1sin4x=sin(4x2x)sin4xsin2x=cot2xcot4x
1sin8x=sin(8x4x)sin8xsin4x=cot4xcot8x
1sin16x=sin(16x8x)sin16xsin8x=cot8xcot16x
nebajcioz

nebajcioz

Beginner2022-02-01Added 15 answers

1sin2x=sinxsin2xsinx=sin(2xx)sin2xsinx=sin2xcosxcos2xsinxsin2xsinx=cotxcot2x
1sin4x=cot2xcot4x
1sin8x=cot4xcot8x
1sin16x=cot8xcot16x
1sin2x+1sin4x+1sin8x+1sin16x=cotxcot16x

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