# 2 math questions

#### cephas1012

Registered Senior Member
two real quick math questions...

1. Is there a trig identity for Sin[x/3] where Sin[x] is known? If so what is it? I am pretty sure I have heard there isnt one...but I could be wrong.

2. This question is a little harder for me to describe. Suppose you start with 1 and say that you can add it together as many times as you like and that you may also add 1*1/2^n together as many times as you like for an integer value of n. So in other words you could get values like 5,4,3/2, 1/2, 1/8, etc. Is there a proof that you can get all rational numbers from this process, or a proof that you cannot? I am really curious about it.

thanks

1. nope

2. how about 1/7?

1. Dunno about the trig thing, can't see any problem there.

2. No 1*1/2^n can't produce all rational numbers, only reals.
I'm not a college-student yet but I guess a proof would be taking the limit when n goes to -infinity and to infinity. Then it can be said that no value x = n will yield a negative result.

I guess...

Originally posted by Redrover
1. nope

2. how about 1/7?

for 1/7th:

Then the goemetric series 1/8+1/8*1/8+1/8*1/8*1/8...to infinity equals 1/8/(1-1/8)=1/7. I didnt actaully say this last time, but I am also allowing myself to add an inifininte number of terms together. That is why I can get this value. I should have made that clear. I would be interested to see if it were possible to get 1/7th without using an infinte series or perhpas 1/3, which also is the sum of an infinite series.

Also, I can subtract terms too. So i can do 1-1-1-1 to get -2.

a little more about this second question i asked. I am pretty sure that if someone could show that you can get all values of 1/k with k being any integer then you could get any rational number just by adding values. I am interested to see if you can get all rationals, or if you can't which values you cant get.

Originally posted by cephas1012
1. Is there a trig identity for Sin[x/3] where Sin[x] is known? If so what is it? I am pretty sure I have heard there isnt one...but I could be wrong.

I think there is one. I just worked this out. Someone check it please, because I am pretty tired.

What I did was convert sin(x/3) to exponential form:

sin(x/3)=(e<sup>ix/3</sup>-e<sup>-ix/3</sup>)/2i

Then I cubed it. Here's the simplified result:

sin<sup>3</sup>(x/3)=(e<sup>ix</sup>-e<sup>-ix</sup>-3e<sup>ix/3</sup>+3e<sup>-ix/3</sup>)/(-8i)

which is equivalent to:

sin<sup>3</sup>(x/3)=(3/4)sin(x/3)-(1/4)sin(x)

or...

3sin(x/3)-4sin<sup>3</sup>(x/3)=sin(x)

So, if you known sin(x) it seems you can solve a cubic equation to find sin(x/3).

sin 3x
= sin(2x+x)
= sin 2x cos x + cos 2x sin x
= 2 sin x cos<sup>2</sup>x + (1 - 2sin<sup>2</sup>x) sin x
= 2 sin x (1 - sin<sup>2</sup>x) + (sin x - 2 sin<sup>3</sup>x)
= 3 sin x - 4 sin<sup>3</sup>x

So

sin x = 3 sin(x/3) - 4 sin<sup>3</sup>(x/3)

Originally posted by James R
sin 3x
= sin(2x+x)
= sin 2x cos x + cos 2x sin x
= 2 sin x cos<sup>2</sup>x + (1 - 2sin<sup>2</sup>x) sin x
= 2 sin x (1 - sin<sup>2</sup>x) + (sin x - 2 sin<sup>3</sup>x)
= 3 sin x - 4 sin<sup>3</sup>x

So

sin x = 3 sin(x/3) - 4 sin<sup>3</sup>(x/3)

Hey, you made that up on the spot!

1/ Is very simple.

Where Sin ( x ) = x - (x^3) /3! + (x^5) / 5! ....

Substitute x for x/3 like so:

sin (x/3) = x/3 - ([x/3]^3) /3! + ([x/3]^5) / 5! ....

Cjwinnit:

He asked for a <b>trig</b> identity. You've given a Taylor series.

Originally posted by James R
Cjwinnit:

He asked for a <b>trig</b> identity. You've given a Taylor series.

Doh!

Ah well, it should be possible to re-arrange the series and end up with a Trig identity but there are easier ways of doing it.

No-one decided to use De Moivre'e Theorem?

cos 3x + i sin 3x = ( cos x + i sin x ) <sup>3</sup>

cos 3x + i sin 3x = cos<sup>3</sup> x + 3 i cos<sup>2</sup> x sin x + 3 cos x i<sup>2</sup> sin<sup>2</sup> x + (i sin x)<sup>3</sup>

then Equate the Imaginary components:

i sin 3x = 3i cos<sup>2</sup> x sin x + (i sin x)<sup>3</sup>

=

sin 3x = 3 (1 - sin<sup>2</sup> x) sin x - (sin x)<sup>3</sup>

=

sin 3x = 3 sin x - 4 sin<sup>3</sup> x

then replace x with x/3:

sin x = 3 sin (x/3) - 4 sin<sup>3</sup> (x/3)

Is that ok?

Last edited:
Just had a thought. Yes, 2/ is correct. It's the number system expressed as base 1/2.

Ok, the Base 10 numbering system:

Express the number 5798. that's 5*1000 + 7*100 + 9*10 + 8*1.

So instead of basing it on the multiples of the numbers 1, 10, 100, 1000..... you base it on 1, 1/2, 1/4, 1/8, ......

You can make any number (including irrationals, but you need an infinite expansion) between 0 and 2 using just those some of those numbers added together, then keep adding 1 until you get the number. Example:

3.7

1+1+1 = 0.7. Now express 0.7 in Base 1/2.

0.7 - 1/2 = 0.2. So 3.7 = 1+1+1+1/2 + 0.2
0.2 - 1/8 = 0.075. So 3.7 = 1+1+1+1/2 + 1/8 + 0.075
0.075 - .....
..
.
See?

Edit, think of it this way.(the number must be positive).

Get the number x and take it's decimal. So in 3.7 the new number is 0.7, which we will call y. if y = 0 then finished.

y is smaller than 1 and larger than 0, so:
• if y is < 1/2 then y = w.
• if y is > 1/2 then remove 1/2 from y to get our new value w. If w = 0 then finished.

w is smaller than 1/2 and larger than 0, so:
• if y is < 1/4 then y = w.
• if y is > 1/4 then remove 1/4 from y to get our new value z. If z = 0 then finished.

etc.....

Last edited: