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Question Number 23243 by paro123 last updated on 27/Oct/17

2^(n−1) sin a×sin 2a×sin 3a×......×sin (n−1)a=n why?

$$\mathrm{2}^{\mathrm{n}−\mathrm{1}} \mathrm{sin}\:\mathrm{a}×\mathrm{sin}\:\mathrm{2a}×\mathrm{sin}\:\mathrm{3a}×......×\mathrm{sin}\:\left(\mathrm{n}−\mathrm{1}\right)\mathrm{a}=\mathrm{n}\:\mathrm{why}? \\ $$

Commented by mrW1 last updated on 28/Oct/17

this statement is wrong!    you can find a lot of examples:  if n=3:  2^(3−1) sin a×sin 2a=4sin a×sin 2a≠3    if a=±π or ±(π/2) or ±(π/3) or ±(π/4).....:  2^(n−1) sin a×sin 2a×sin 3a×......×sin (n−1)a=0≠n

$$\mathrm{this}\:\mathrm{statement}\:\mathrm{is}\:\mathrm{wrong}! \\ $$$$ \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{find}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{of}\:\mathrm{examples}: \\ $$$$\mathrm{if}\:\mathrm{n}=\mathrm{3}: \\ $$$$\mathrm{2}^{\mathrm{3}−\mathrm{1}} \mathrm{sin}\:\mathrm{a}×\mathrm{sin}\:\mathrm{2a}=\mathrm{4sin}\:\mathrm{a}×\mathrm{sin}\:\mathrm{2a}\neq\mathrm{3} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{a}=\pm\pi\:\mathrm{or}\:\pm\frac{\pi}{\mathrm{2}}\:\mathrm{or}\:\pm\frac{\pi}{\mathrm{3}}\:\mathrm{or}\:\pm\frac{\pi}{\mathrm{4}}.....: \\ $$$$\mathrm{2}^{\mathrm{n}−\mathrm{1}} \mathrm{sin}\:\mathrm{a}×\mathrm{sin}\:\mathrm{2a}×\mathrm{sin}\:\mathrm{3a}×......×\mathrm{sin}\:\left(\mathrm{n}−\mathrm{1}\right)\mathrm{a}=\mathrm{0}\neq\mathrm{n} \\ $$

Commented by paro123 last updated on 28/Oct/17

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