Question

17. Show that sin A + sin (120° + A) + sin (120° - A) = 0.​

Answer 1

Answer:

math]\sin\, \alpha +\sin\,\beta = 2 \sin \dfrac{\alpha + \beta}{2} \cos\dfrac{\alpha - \beta}{2}[/math]

[math]\implies \sin\,A + \sin (120^\circ + A) + \sin (A - 120^\circ)[/math]

[math]= \sin\,A + 2 \sin \dfrac{(120^\circ + A + A - 120^\circ)}{2}\cos \dfrac{\{120^\circ + A - (A - 120^\circ)\}}{2}[/math]

[math]= \sin\,A + 2 \sin\,A \cos\,120^\circ[/math]

[math]= \sin\,A + 2 \sin\,A\left( \dfrac{-1}{2}\right)[/math]

[math]= \sin\,A - \sin\,A = 0[/math]

Step-by-step explanation:

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Answer 2

Answer:

sin $A$−sin $A = 0$

Step-by-step explanation:

sin $a$ +sin β =2sin $\frac{a+b}{2}$ cos $\frac{a-b}{2}$

sin $A$  $+$ sin$(120^ o+A)$ +sin$(A-120^o)$

=sin $A$+2sin$\frac{120^o + A + A - 120^o}{2}$cos$(120^o + A - A (A - 120^o))$

=sin $A + 2$ sin A cos $120^o$

=sin $A$+2 sin A $(\frac{-1}{2})$

=sin $A$−sin A=0