Grade 11Trigonometry

Write the value of sin10+sin20+sin30+....+sin360.Show me the solution plz

Nimish Singh

13 Years agoGrade 11

5 Answers

Shaurya Gupta

13 Years ago

= sin(10+36x5)sin(37x5)/ sin5

= -sin10.-sin5/sin5

= sin10

eq 1 is a formula

sin(a)+sin(a+b)+sin(a+2b)............sin(a+(n-1)b) = sin(a+(n-1)b/2)sin(nb/2)/ sin(b/2)

lokesh soni

13 Years ago

sin10+sin20+sin30+....+sin360=0

sin10+sin20+sin30+sin40+...sin180+sin(360-170)+......+sin(360-40)+sin(360-30)+sin(360-20)+sin360-10)+sin360

sin360-x=-sinx

hence all terms cancel and

sin10+sin20+sin30+....+sin360=0

Dharm Prakash

9 Years ago

a = 10°, b = 10°, then the value of n is a+(n-1)b = 360°=> 10° + (n-1) × 10° = 360°=> (n-1) × 10° = 360 - 10°=> (n-1) × 10° = 350°=> (n-1) = 350°/10°=> (n-1) = 35°=> n = 36°Sin(360°/2).Sin{(36×10)/2}/Sin(10/2) = Sin(180°).Sin(180°)/Sin(5°) = 0

Amrita Kashyap

7 Years ago

Sin10⁰+ sin20⁰+ sin30⁰+.......+ sin360⁰

=sin(360-350)⁰+ sin(360-340)⁰+ sin(360-330)⁰+......+sin350⁰+ sin360⁰

= - sin350⁰- sin340⁰-sin330⁰+......+ sin330⁰+sin340⁰+ sin350⁰+ sin360⁰

= sin180⁰+sin360⁰

^{= 0}

Rishi Sharma

5 Years ago

Dear Student,
Please find below the solution to your problem.

sin10+sin20+sin30+....+sin360=0
sin10+sin20+sin30+sin40+...sin180+sin(360-170)+......+sin(360-40)+sin(360-30)+sin(360-20)+sin360-10)+sin360
sin360-x=-sinx
hence all terms cancel and
sin10+sin20+sin30+....+sin360=0

Thanks and Regards