Question

Calculati sin(1)+sin(2)+...sin(90)=?

Answer 1

S = sin(1)+sin(2)+...+sin(89)+sin(90)

S = sin(1)+sin(2)+...+sin(89)+1

S = cos(89)+cos(88)+...+cos(1)+1

(S-1) = cos(1)+cos(2)+...+cos(89)

(S-1)•sin(1/2) =

= sin(1/2)•cos(1) + sin(1/2)•cos(2) + ... + sin(1/2)•cos(89)

Aplic formula:

sin(x)•cos(y) = {sin[(x+y)/2] - sin[(x-y)/2]}/2

(S-1)•sin(1/2) = [sin(3/2) - sin(1/2)]/2 + [sin(5/2) - sin(3/2)]/2 + [sin(7/2) - sin(5/2)]/2 + ... + [sin(179/2) - sin(177/2)]/2

Aceasta este o sumă telescopică în care se reduc aproape toți termenii, iar rezultatul este:

= [sin(179/2) - sin(1/2)]/2

(S-1) = [sin(179/2) - sin(1/2)]/[2sin(1/2)]

= (1/2)•[sin(179/2)]/[sin(1/2)] - 1/2

= (1/2)• [sin(179/2)]/[cos(90 - 1/2)] - 1/2

= (1/2)•[sin(179/2)]/[cos(179/2)] - 1/2

= (1/2)•tg(179/2) - 1/2

S = (1/2)•tg(179/2) - 1/2 + 1

⇒ S = (1/2)•tg(179/2) + 1/2