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Question Number 13305 by prakash jain last updated on 18/May/17
Does  sin A=sin B  imply sin (A/2)=sin(B/2)?  For example A=π and B=−π?
$$\mathrm{Does} \\ $$$$\mathrm{sin}\:\mathrm{A}=\mathrm{sin}\:\mathrm{B} \\ $$$$\mathrm{imply}\:\mathrm{sin}\:\frac{\mathrm{A}}{\mathrm{2}}=\mathrm{sin}\frac{\mathrm{B}}{\mathrm{2}}? \\ $$$$\mathrm{For}\:\mathrm{example}\:\mathrm{A}=\pi\:\mathrm{and}\:\mathrm{B}=−\pi? \\ $$
Answered by ajfour last updated on 18/May/17
A=nπ+(−1)^n B  when sin A=sin B ;  sin (A/2)=sin (((nπ)/2)+(−1)^n (B/2))  for B=−π ; A=π   we have n=2  sin (A/2)=sin (((2π)/2)+(−1)^2 (B/2))  sin (A/2) =sin (π+(B/2)) ≠ sin (B/2) .
$${A}={n}\pi+\left(−\mathrm{1}\right)^{{n}} {B} \\ $$$${when}\:\mathrm{sin}\:{A}=\mathrm{sin}\:{B}\:; \\ $$$$\mathrm{sin}\:\frac{{A}}{\mathrm{2}}=\mathrm{sin}\:\left(\frac{{n}\pi}{\mathrm{2}}+\left(−\mathrm{1}\right)^{{n}} \frac{{B}}{\mathrm{2}}\right) \\ $$$${for}\:{B}=−\pi\:;\:{A}=\pi\:\:\:{we}\:{have}\:{n}=\mathrm{2} \\ $$$$\mathrm{sin}\:\frac{{A}}{\mathrm{2}}=\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{2}}+\left(−\mathrm{1}\right)^{\mathrm{2}} \frac{\boldsymbol{{B}}}{\mathrm{2}}\right) \\ $$$$\mathrm{sin}\:\frac{{A}}{\mathrm{2}}\:=\mathrm{sin}\:\left(\pi+\frac{\boldsymbol{{B}}}{\mathrm{2}}\right)\:\neq\:\mathrm{sin}\:\frac{\boldsymbol{{B}}}{\mathrm{2}}\:. \\ $$

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