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Question Number 160762 by mathlove last updated on 06/Dec/21

sin 10+sin 20+sin 30+sin 40+∙∙∙∙+sin 360=?

$$\mathrm{sin}\:\mathrm{10}+\mathrm{sin}\:\mathrm{20}+\mathrm{sin}\:\mathrm{30}+\mathrm{sin}\:\mathrm{40}+\centerdot\centerdot\centerdot\centerdot+\mathrm{sin}\:\mathrm{360}=? \\ $$

Commented by som(math1967) last updated on 06/Dec/21

0

$$\mathrm{0} \\ $$

Commented by mathlove last updated on 06/Dec/21

How  solution

$${How}\:\:{solution} \\ $$

Commented by som(math1967) last updated on 06/Dec/21

sinx+sin2x+...+sinnx=((sin(((n+1)/2))xsin((nx)/2))/(sin(x/2)))  if x=10   ∴10,20,30...360     36 term  ∴n=36  ∴sin10+sin20+...+sin360  =((sin((370)/2)sin((360)/2))/(sin((10)/2)))  =((sin185sin180)/(sin5))=0  [∵sin180=0]

$${sinx}+{sin}\mathrm{2}{x}+...+{sinnx}=\frac{{sin}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right){xsin}\frac{{nx}}{\mathrm{2}}}{{sin}\frac{{x}}{\mathrm{2}}} \\ $$$${if}\:{x}=\mathrm{10}\: \\ $$$$\therefore\mathrm{10},\mathrm{20},\mathrm{30}...\mathrm{360}\:\:\:\:\:\mathrm{36}\:{term} \\ $$$$\therefore{n}=\mathrm{36} \\ $$$$\therefore{sin}\mathrm{10}+{sin}\mathrm{20}+...+{sin}\mathrm{360} \\ $$$$=\frac{{sin}\frac{\mathrm{370}}{\mathrm{2}}{sin}\frac{\mathrm{360}}{\mathrm{2}}}{{sin}\frac{\mathrm{10}}{\mathrm{2}}} \\ $$$$=\frac{{sin}\mathrm{185}{sin}\mathrm{180}}{{sin}\mathrm{5}}=\mathrm{0} \\ $$$$\left[\because{sin}\mathrm{180}=\mathrm{0}\right] \\ $$

Commented by MJS_new last updated on 06/Dec/21

sin A° =−sin (180+A)°  we don′t need to know more

$$\mathrm{sin}\:{A}°\:=−\mathrm{sin}\:\left(\mathrm{180}+{A}\right)° \\ $$$$\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{need}\:\mathrm{to}\:\mathrm{know}\:\mathrm{more} \\ $$

Commented by som(math1967) last updated on 06/Dec/21

yes sir  sin10+sin20+...+sin180  +sin190+sin200+...sin360  =sin10+sin20+...+sin180  −sin10−sin20...+sin360  =0

$${yes}\:{sir} \\ $$$${sin}\mathrm{10}+{sin}\mathrm{20}+...+{sin}\mathrm{180} \\ $$$$+{sin}\mathrm{190}+{sin}\mathrm{200}+...{sin}\mathrm{360} \\ $$$$={sin}\mathrm{10}+{sin}\mathrm{20}+...+{sin}\mathrm{180} \\ $$$$−{sin}\mathrm{10}−{sin}\mathrm{20}...+{sin}\mathrm{360} \\ $$$$=\mathrm{0} \\ $$

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