Proving the trigonometric identity

1. Jul 4, 2012

justwild

1. The problem statement, all variables and given/known data

To prove that $\sum$ over m=1 to 15 of sin(4m-2) = 1/4sin2, where all angles are in degress

2. Relevant equations

3. The attempt at a solution
Tried to solve it using identity sinx+siny=2sin((x+y)/2)cos((x-y)/2)..but all attempts failed..help

2. Jul 4, 2012

Saitama

Try opening the sigma first.
Assume that the sum is equal to S.
$$S=sin(2)+sin(6)+sin(10)+sin(14)+......$$
Can you write it in the form:
$$S=sin(a)+sin(a+d)+sin(a+2d).....$$
where d is the difference between two consecutive angles.

Last edited: Jul 4, 2012
3. Jul 4, 2012

Curious3141

Are you allowed to use complex numbers? There's a way to do this using complex numbers, the sum of a geometric series, followed by a couple of trig identities, and it's not too hard.

4. Jul 4, 2012

Saitama

Complex numbers? I would like to know that. :tongue2:
I have derived a formula for this series of sine (when the angles are in Arithmetic Progression) but it's really lengthy. Please post your method of Complex numbers if the OP doesn't reply. That would help me.

5. Jul 4, 2012

Yukoel

Hello pranav,
I think what Curious3141 implies is that you express sin(θ) as the difference of a complex number and its conjugate in Euler form.The various terms form a G.P. which can be summed up and the rest of it is a little of trigonometry.
Hoping this helps.

regards
Yukoel

6. Jul 4, 2012

amiras

You was on the right track, just group numbers this time and do some trick:

Your sum expanded will look something like this:
sin2 + sin6 + sin10 + ... + sin50 + sin54 + sin 58

Now regroup them:
(sin2 + sin58) + (sin6 + sin54) + (sin10 + sin40) + (... groups)+ sin 30

Now notice that the all brackets have average of angle 30. (2+58, 6+54...)

That would be a good idea to define the angles from this average center 30, so instead lets write: sin2 = sin(30-28) and sin58 = sin(30+28)... etc.

Now we got something like this:

[ sin(30-28) + sin(30+28) ] + [ sin(30-24) + sin(30+24) ] + ... + sin30

For group members in brackets we can apply your suggested formula:
sin(a-x)+sin(a+x) = 2sina cosx

Now series simplifies to:
2sin30(cos28 + cos24 + cos 20 + cos16 + cos12 + cos8 + cos4) + sin30

Since sin30 = 1/2

Our series now is: cos28+cos24+cos20+...+ 1/2

Can you take it from here?

7. Jul 4, 2012

justwild

Yes you can use complex numbers..

8. Jul 4, 2012

justwild

Finally found the solution...

Here it goes,
let x=sin2+sin6+....+sin30+...sin58
multiplying 2sin2 on both sides,
2(sin2)x=2sin$^{2}$2+2sin2(sin6)+2sin2(sin10)...2sin2(sin58)
simplifying further,
2(sin2)x=1+cos4+cos8-cos4+cos12-cos8......cos60-cos56
this implies,
2(sin2)x=1-cos4+cos4+cos8-cos8+cos12-cos12+....cos60
2(sin2)x=1-cos60
further,
2(sin2)x=1/2
Therefore,
x=1/4sin2
Which is the required solution.

Actually this question came in an entrance exam (IIT-JEE) in 2009--Q27 Maths section..you can download question paper at http://www.jee.iitb.ac.in/images/2009p1.pdf

9. Jul 4, 2012

justwild

10. Jul 4, 2012

Curious3141

Sure. What you came up with is a very neat proof. But it looks a little "tailor-made" (post hoc) to the question. By which I mean that you chose to multiply by 2sin2 because of that term in the RHS.

Here's a slightly longer, but more general, method:

Start by recognising that your series S is composed of sines of arguments in arithmetic progression. Let the first term be "a" and the common difference be "d". Pranav has posted this form above.

This sum can also be represented as the Imaginary part of this complex series i.e. S = Im(z), where:

$$z = e^{ia} + e^{i(a + d)} + ... + e^{i[a + (n-1)d]}$$

which is in fact a geometric series. Sum it the usual way.

$$z = \frac{e^{ia}(e^{ind} - 1)}{e^{id} - 1}$$

Multiply both top and bottom by the complex conjugate of the denominator ("realise the denominator"), group terms, simplify:

$$z = \frac{e^{i[a+(n-1)d]} - e^{[i(a+nd)]} - e^{i(a-d)} + e^{ia}}{2(1 - \cos d)}$$

from which you extract the imaginary part:

$$S = \frac{\sin{[a+(n-1)d]} - \sin{[(a+nd)]} - \sin{(a-d)} + \sin{a}}{2(1 - \cos d)}$$

This is a general formula that allows you to compute the sum of a series of sines of arguments in AP. If you replace the sin with cos in the formula, you'll get the same for sums of cosines. The formula can likely be simplified further, but I just applied it in that form for this question.

Now simply plug in a = 2 degrees, d = 4 degrees. Technically, in analysis, we generally work in radian measure, but since we didn't do any differentiation or integration, the choice of measure doesn't matter here, and the formula still holds.

So:

$$S = \frac{\sin 58 - \sin 62 - \sin{(-2)} + \sin 2}{2(1 - \cos 4)}$$

apply factor formula (i.e. $\sin A - \sin B = 2\sin{\frac{A-B}{2}}\cos{\frac{A+B}{2}}$) to the first two terms and simplify:

$$S = \frac{-2\sin 2\cos 60 + 2\sin 2}{2(1 - \cos 4)} = \frac{\sin 2}{2(1 - \cos 4)}$$

Finally, apply half-angle (or double-angle) formula to the denominator:

$$S = \frac{\sin 2}{2(1 - 1 + 2\sin^2 2)} = \frac{\sin 2}{4\sin^2 2} = \frac{1}{4\sin 2}$$

as required.

Last edited: Jul 4, 2012
11. Jul 4, 2012

justwild

Great technique. I didn't know that we can even do operations by taking the imaginary part of euler's formula this way. Thanks.

12. Jul 4, 2012

Saitama

Hey justwild, seems like you got a lot of help since i went offline. Here's a general formula for the series of sine when the angles are in AP.
$$S=\frac{\sin\frac{nd}{2}}{\sin\frac{d}{2}}\cdot sin\frac{2a+(n-1)d}{2}$$

And thank you Curious for the alternative method!

13. Jul 4, 2012

Curious3141

After simplifying my expression, I get the same form, although I prefer to express it as:

$$S_{\sin} = \frac{\sin{[a + \frac{1}{2}(n-1)d]}\sin{(\frac{1}{2}nd)}}{\sin{(\frac{1}{2}d)}}$$

and the analogous expression for a sum of cosines is:

$$S_{\cos} = \frac{\cos{[a + \frac{1}{2}(n-1)d]}\sin{(\frac{1}{2}nd)}}{\sin{(\frac{1}{2}d)}}$$