Question

sinA+sin2A+sin3A+....+sin(nA)?

Answer 1

Write it backwards and add them together....

Answer 2

:/

Answer 3

n/2 ( Sin A + Sin nA)

Answer 4

Why?

Answer 5

Write it backwards and add them together....

Answer 6

Ok, thanks.

Answer 7

i do not think so ,sina +sin(na) is not equal to sin2a+sin{( n-1)a} can not calculate that way

Answer 8

So far I tried to generalize using the identities:

sin (x + b) = sinxcosb + sinbcosx
cos (x +b) = cosxcosb - sinxsinb

And I got:


sinA+sin2A+sin3A+....+sin(​nA)

sin2A = sinAcosA + sinAcosA
sin2A = 2 sinAcosA

sin3A = (2A + A) = sin2A cosA + sinAcos2A
2 sinAcosA*cosA + sinAcos2A
2 sinAcosA*cosA + sinAcos(A+A)
2 sinAcosA*cosA + sinA ( cosA cosA - sinA sinA)
2 sinAcosA*cosA + sinA ( cos^2 A - sin^2A)
2 sinAcos^2 A+ sinA ( cos^2 A - sin^2A)

sin^2 A + cos^2 A = 1
sin^2 A = 1 - cos^2 A

2 sinAcos^2 A+ sinA ( cos^2 A - ( 1 - cos^2 A) )
2 sinAcos^2 A+ sinA ( cos^2 A -1 + cos^2 A )

2 sinAcos^2 A+ sinA ( 2 cos^2 A -1 )
2 sinAcos^2 A+ 2 sinA cos^2 A - sinA
4 sin A cos^2 A - sin A

sin2A = 2 sinAcosA
sin3A = 4 sin A cos^2 A - sin A

sin 4A = sin (3A + A) = sin 3A cos A + sin A cos 3A

(4 sin A cos^2 A - sin A) cos A + sin A cos 3A
(4 sin A cos^2 A - sin A) cos A + sin A cos (2A + A)
(4 sin A cos^2 A - sin A) cos A + sin A ( cos 2A cos A - sin 2A sin A)
(4 sin A cos^2 A - sin A) cos A + sin A ( (2 cos^2 A -1) cos A - sin 2A sin A)

Answer 9

‎(4 sin A cos^2 A - sin A) cos A + sin A ( (2 cos^2 A -1) cos A - (2 sinAcosA) sin A)

4 sin A cos^3 A - sin A cos A + 2 sin A cos^3 A -sin A cos A - 2sin^3A cosA

6 sin A cos^3 A - 2 sin A cos A - 2sin^3A cosA
2 sin A cos A ( 3 cos^2 A - 1 - sin^2 A)
2 sin A cos A ( 3 cos^2 A - 1 - (1 - cos^2 A))
2 sin A cos A ( 3 cos^2 A - 1 - 1 + cos^2 A)
2 sin A cos A ( 4 cos^2 A - 2)
8 sin A cos^3 A - 4 sin A cos A

sin 4A = 8 sin A cos^3 A - 4 sin A cos A

sin A = sin A
sin 2A = 2 sinAcosA
sin 3A = 4 sin A cos^2 A - sin A
sin 4A = 8 sin A cos^3 A - 4 sin A cos A

Answer 10

I just thought I made a booboo here.....:-(

Answer 11

sin A + sin 2A = sin A (1 + 2 cos A)
sin A + sin 2 A + sin 3A = sin A + 2 sinA cos A + 4 sin A cos^2 A - sin A

sin A ( 1 + 2 cosA + 4 cos^2 A - 1) =
sin A (2 cosA + 4cos^2 A) =
2 sinA cos A (1 + 2 cos A)

sin A + sin 2A + sin3A + sin 4A =
sinA cos A (2 + 4 cos A) + 8 sin A cos^3 A - 4 sin A cos A
2 sinA cos A (1 + 2 cos A) + 2 sin A cos A ( 4 cos^2 A - 2)
(2 sinAcosA) ( 1 + 2cosA + 4 cos^2 A -2)
(2 sinAcosA) ( 2cosA + 4 cos^2 A -1)

Answer 12

maybe we need the taylor formula

Answer 13

I think I'll give up. ://

Answer 14

No, no...have to split in odds and evens...

Answer 15

:o

Answer 16

How?

Answer 17

Give me a while, not a machine..:-)

Answer 18

Alright.

Answer 19

i think this question must belong to infinite series (calcuslu ii)

Answer 20

Ok, u have a look at that, maybe ur right (but this is finite)

For the even part (2A,4A etc) I have

n(n+2)/2 SinA CosA (somebody check).

Let me see if I can get something out of the odd parts

Answer 21

This is a right pain, you have 2 cases (n even or odd)

The even components are OK (above) but the odd are more trouble.

In the end I decided to see what Wolfram had to say, I got

Sin(nA/2) Sin ((n+1)A/2) / Sin(A/2)

I think there is some way to do this with complex numbers, not sure, though...

Answer 22

Actually, maybe that should simplify to just

Sin ((n+1)A/2)

I see the way to deal with the odd even u need half angle so multiply the whole shebang by sin (A/2) and use

sin x sin y = 1/2 (cos(x-y) -cos(x+y)) and then it just frops out:-):-)

Answer 23

Drops out I mean...

Answer 24

Because all the half angles just cancel each other out...

Answer 25

\[\sum_{k=1}^{n}\sin(k A) = \frac{\sin\left(\frac{n A}{2}\right) \sin\left(\frac{ (n+1) A}{2}\right)}{\sin\left(\frac{A}{2}\right)}\]

Can be proven by indiction

Answer 26

Yes, that's same as Wolfram, u don't need to do that....

Answer 27

okay I got this \[SinA + Sin B = 2Sin \frac{A+B}{2} Cos\frac{A -B}{2}\]
If we do this with odd terms we get even sin terms

Answer 28

True, but the way I just said is dead easy, about 3 or 4 lines is all and u get

Sin ((n+1)A/2)

Answer 29

Oh yeah sorry i didn't read that just started replying

Answer 30

Well estudier you talk about complex number way of solving it

Answer 31

not talk about but mentioned it

Answer 32

what if we take sin a to cos (pi/2 - a)

Answer 33

then imaginary is zero

Answer 34

and pi/2 -a = npi

Answer 35

then the whole angle becomes pi x n ..then maybe polar form e^i pi n now n should be integers

Answer 36

ah i don't know anyway your method is better thanks for that

Answer 37

Sorry, phone call...I will have a look at that...

Answer 38

: ) thanks

Answer 39

U can use the complex equivalent of half angle set up.

If you multiply series by i and add it to Cos A + Cos 2A etc u get

a complex sum of form a + ib then use Euler formula to get

cos A = 1/2( e^iA + e^iA) and sin x = 1/2 (e^iA - e^iA) ->

e^iA + e^2iA + etc which is geometric series and then you just compare real and imaginary terms it comes out to the Wolfram answer I put up before (I made a silly cancelling error in the trig way I did it).

Still think the trig way is more straightforward once u see the half angle trick.