Question

Simplify /compute
sin(0.7t) + cos(1.1t) +sin(1.5t) + cos(1.9t) + sin(2.3t) + cos(2.7t) + sin(3.1t) + cos(3.5t) +sin(3.9t) =

Answer 1

try grouping like this sin(0.7t) + sin(2.3t) + cos(1.1t) + cos(1.9t) + sin(1.5t) + cos(3.5t) + sin(3.1t) + sin(3.9t) + cos(2.7t)
:P

Answer 2

Yes, go on please.

Answer 3

ah.. leave it :P

Answer 4

It will be solved eventually.

Answer 5

it maybe solved :P not sure by my side

Answer 6

I give you hope. (and change ?...)

Answer 7

@mathslover welcome help

Answer 8

Well, no change so far

Answer 9

Hey there @mathslover you multiplied (and grown in NUMBER !) congrats

Answer 10

Any ideas people ?

Answer 11

I wouldn't ask this question unless it was educational and elegant

Answer 12

No luck bro :P

Answer 13

Well what about this idea @vikrantg4 @mathslover
\[\cos(\alpha) = 0.5(e^{i \alpha} + e^{-i \alpha})\]

Answer 14

omg i don't know details in complex numbers lool

Answer 15

\[\sin(\beta) = -i 0.5(e^{i\beta} - e^{-i\beta})\]

Answer 16

Forget complex numbers in general use this formula only

Answer 17

it's gonna be a pain to use this in 9 expresions

Answer 18

The whole point is IT IS NOT PAIN there is a way

Answer 19

lool :PP

Answer 20

@zzr0ck3r I gave a guidance to solution

Answer 21

@vikrantg4 Try using your own first advice and these

Answer 22

excuse me?

Answer 23

I thought you might help @zzr0ck3r

Answer 24

Ok time to spoil or not yet ?

Answer 25

It' getting more complex using my method

Answer 26

Replace everything with exponents - then what do you get ?

Answer 27

Lets say\[e^{i\alpha} + e^{i 2\alpha} = ?\]

Answer 28

Hi @ajprincess

Answer 29

Thanks Flute-passionate @mathslover

Answer 30

.welcome. ;)

Answer 31

Well ladies and gentleman - what is this\[e^\sigma + e^{2\sigma} + e^{3\sigma} + e^{4\sigma}\]

Answer 32

It is a .....it is Ahhh ....

Answer 33

expression. For sure..

Answer 34

geometric sequence

Answer 35

;\

Answer 36

Great @Coolsector , NOW ?

Answer 37

q=e^a

Answer 38

Yes but what do we DO now ?

Answer 39

using the formula for sum

Answer 40

Step forward - what will it be in this case ?

Answer 41

thx @Coolsector

Answer 42

well this is great.. thank you for posting this

Answer 43

how can i do it ? :| lol i have no idea

Answer 44

http://www.wolframalpha.com/

Answer 45

Hi @sauravshakya i must admit, pleased to see you here

Answer 46

Thanx and GOOD ONE!

Answer 47

In fact there are A LOT of practical applications of this
1) Z-transform
2) Digital Signal Processing filters design
3) Complex Integration
.
.
.

Answer 48

Thanks @TheViper pal !

Answer 49

\(\Huge{\color{violet}{\ddot{\smile}}}\)

Answer 50

You tell me in PM how one does THAT !

Answer 51

PM ?

Answer 52

Private messg.

Answer 53

\[ \Huge{\color{green}{\ddot{\smile}}} \]

Answer 54

Mikael
4
Finding the sum of a trigonometric series uses De Moivre's Theorem, geometric progression, trigonometric identities, and complex numbers. Top universities for mathematics test prospective students by setting questions in their entrance tests that stretch the most capable students. The example used here has been used in the past by Cambridge University and other top math colleges. The question is an excellent one, that uses De Moivre's Theorem, geometric progressions, trigonometric identities, complex numbers, and algebraic manipulation. Problem: Find the General Solution to cos(x) + cos(2x) + cos(3x) + ... + cos(nx) The core of the solution is to recognise that cos.x + cos.2x + cos.3x + ... + cos.nx = Real part of (cos.x + i.sin.x + cos.2x + i.sin.2x + cos.3x + i.sin.3x + ... + cos.nx + i.sin.nx) (where i = √-1) = Real part of [ (cos.x + i.sin.x) + (cos.x + i.sin.x)^2 + (cos.x + i.sin.x)^3 + ... + (cos.x + i.sin.x)^n ] ... [Eq'n 1] This is a geometric progression: let Y = (cos.x + i.sin.x), and the series becomes Real part of (Y + Y^2 + Y^3 + ... + Y^n) From the rules of a geometric progression, this is = Real part of [Y.(Y^n - 1) / (Y - 1)] Now substitute Y = (cos.x + i.sin.x) back in again, and the sum becomes = Real part of (cos.x + i.sin.x).[ (cos.x + i.sin.x)^n –1] / (cos.x + i.sin.x - 1) = Real part of (cos.x + i.sin.x).[ (cos.nx + i.sin.nx) –1] / (cos.x + i.sin.x - 1) Rationalise The Denominator The denominator is complex. To get rid of the “i” term, multiply numerator and denominator by the complex conjugate of the denominator, (cos.x –1 – i.sin.x), and the summation becomes =(cos.x + i.sin.x).[ (cos.nx + i.sin.nx) –1].(cos.x -1 + i.sin.x ) / [ (cos.x -1 + i.sin.x ).( cos.x -1 - i.sin.x) ] For clarity, the denominator will be expanded first: = (cos.x -1 + i.sin.x ).( cos.x -1 - i.sin.x) This is simply the difference of two squares: = (cos.x – 1)^2 – (i.sin.x)^2 = (cos.x)^2 – 2.cos.x. + 1 – (i^2.(sin.x)^2) = (cos.x)^2 – 2.cos.x. + 1 + (sin.x)^2 But (cos.x)^2 + (sin.x)^2 = 1, so the denominator becomes = 2 – 2.cos.x =2.(1 – cos.x) Evaluate the Numerator The numerator is = (cos.x + i.sin.x).[ (cos.nx + i.sin.nx) –1].( cos.x -1 - i.sin.x) = (cos.x.cos.nx + cos.x.i.sin.nx - cos.x + i.sin.x.cos.nx + i.sin.x.i.sin.nx - i.sin.x)).( cos.x -1 - i.sin.x ) But from the standard trigonometric identities, (cos.x.cos.nx - sin.x.sin.nx) = cos(nx + x) = cos(n+1).x and cos.x.i.sin.nx + i.sin.x.cos.nx = sin(nx + x) = sin(n+1).x So the numerator becomes = (cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x). ( cos.x - 1 - i.sin.x ) = cos.x.(cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x) - (cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x) - i.sin.x.(cos(n+1).x + i.sin(n+1).x - cos.x - i.sin.x) = cos.x.cos(n+1).x + i.cos.x.sin(n+1).x - cos².x - i.cos.x.sin.x - cos(n+1).x - i.sin(n+1).x + cos.x + i.sin.x - i.sin.x.cos(n+1).x - i².sin.x.sin(n+1).x + i.sin.x.cos.x + i².sin.x.sin.x = cos.x.cos(n+1).x + i.cos.x.sin(n+1).x - cos².x - i.cos.x.sin.x - cos(n+1).x - i.sin(n+1).x + cos.x + i.sin.x - i.sin.x.cos(n+1).x + sin.x.sin(n+1).x + i.sin.x.cos.x - sin².x ... [Eq'n 2] = cos.x + cos.2x + ... cos.nx + i.(sin.x + sin.2x + ... + sin.nx) Find cos.x + cos.2x + cos.3x + ... + cos.nx From Equation [1], the sum of the series is the real part of the geometric progression: Gathering real terms, the real part of equation 2 = cos.x.cos(n+1).x - cos².x - cos(n+1).x + cos.x + sin.x.sin(n+1).x - sin².x Simplifying, cos.x.cos(n+1).x + sin.x.sin(n+1).x = cos[(n+1).x - x] = cos.nx, and -cos²x - sin²x = -(cos²x + sin²x) = -1, so the real part becomes = cos.nx - cos.(n+1).x - 1 + cos.x So the sum of cos.x + cos.2x + ... + cos.nx = numerator / denominator = (cos.nx - cos.(n+1).x - 1 + cos.x) / 2.(1 - cos.x) = (cos.nx - cos.(n+1).x) / (2 - 2.cos.x) - 1 / 2 Q.E.D. To verify this, choose x = π / 5 and n = 3 as an example: (cos.3.π / 5 - cos.(4.π / 5)) / (2 - 2.cos.π / 5) - 1 / 2 = 0.809 cos(π / 5) + cos(2.π / 5) + cos(3.π / 5) = 0.809 + 0.309 - 0.309 = 0.809 Sum of Trigonometric Series Using De Moivres Theorem: cosx+cos2x+ | Suite101.com http://suite101.com/article/sum-of-trigonometric-series-using-de-moivres-theorem-cosxcos2x-a369345#ixzz27BKBaewU

Answer 55

cos(3.5t) +sin(3.9t)
cos(3.9t-0.4t) +sin(3.9t)
cos(3.9t) cos(0.4t) +sin(3.9t) sin(0.4t) +sin(3.9t)
Like this you can reduce all the terms to sin/cos(3.9t) multiplied by a constant.

Answer 56

Well @henpen you are correct (in your general hunch) but guess what - the most elegant and short way to prove and UNDERSTAND addition formulas is BY De Moivre theorem !

Answer 57

I do understand that, just was too lazy to read the full proof.

Answer 58

Oddly enough I encountered de Moivre's theorem before any addition formulae. Anyway, thanks for bringing me here again- the proof is very elegant!

Answer 59

Hey You sooth my tormented soul. This place is very double edged regarding how people react to real elegant things shown them....
For example this stinking sid whatever http://openstudy.com/study#/updates/505e1877e4b02e1394101626