Math help? c:

Jabatheblob1 · 2015-03-20 05:58:43

2sin(5x) = 1-cos(10x) Prove it's true please if anyone can, struggling

Slushie · 2015-03-20 06:13:51

Use the double angle identity for cos:
cos(2x)=1-2sin^2(x) => cos(10x)=1-2sin^2(5x)

Plug that into the given equation:

2sin(5x)=1-(1-2sin^2(5x))
2sin(5x)=2sin^2(5x)
sin^2(5x)-sin(5x)=0
sin(5x)(sin(5x)-1)=0
sin(5x)=0 or sin(5x)=1
5x=pi*n or 5x=pi/2+2pi*n

x=(pi*n)/5 or x=pi/10+(2pi*n)/5

Jabatheblob1 · 2015-03-20 07:04:22

I don't want X, i want to prove that 2sin(5x) = 1-cos(10x)

Stagecrew · 2015-03-20 11:56:14

That statement isn't always true, it is only true for the set of solutions yes found. They can't be equivalent because cos(x) and sin(x) produce results ranging from -1 to 1, so 1-cos(10x) can be at least 0 and at most 2. Using the same reasoning, 2sin(5x) is at least -2 and at most 2. This means the functions are not the same.

rgl32

Lionhart · 2015-03-20 17:13:20

Jabatheblob1 wrote:

2sin(5x) = 1-cos(10x) Prove it's true please if anyone can, struggling

Unless you copied the problem wrong, there is no proof. Just look at the graphs. Completely different.

The closest I can get is 2sin(5x)=-2sin^2(5x).

Sybil · 2015-03-20 21:57:51

i don't know why you're getting help from the internet, especially here lmao

Twipply · 2015-03-20 23:21:34

Sybil wrote:

i don't know why you're getting help from the internet, especially here lmao

Maybe because a vast number of people online are very, very good at maths and can answer this with ease? Even some of them on here.

Pingohits

Jabatheblob1 · 2015-03-20 23:54:49

Many of them are on here, and yes i did write the question wrong, i went in and talked to my math teacher. It's sin^2(5x) = 1-cos(10x)

Pingohits · 2015-03-21 02:50:58

also half of this forum hasn't even touched a single trig problem yet

Lionhart · 2015-03-21 06:59:43

Jabatheblob1 wrote:

Many of them are on here, and yes i did write the question wrong, i went in and talked to my math teacher. It's sin^2(5x) = 1-cos(10x)

In that case, it's just as simple as using the Double Angle Formula for cosine. I forgot the parentheses last time so that's how I ended up with a negative answer. Here is a possible correct answer. (There are multiple ways to prove something is true.)

1) 2sin^2(5x) = 1 - cos(10x)
2)            = 1 - cos(2 * 5x)
3)            = 1 - (1 - 2sin^2(5x))
4)            = 1 - 1 + 2sin^2(5x)
5) 2sin^2(5x) = 2sin^2(5x)

rubixguy · 2015-03-21 21:06:09

Lionhart wrote:

Jabatheblob1 wrote:
Many of them are on here, and yes i did write the question wrong, i went in and talked to my math teacher. It's sin^2(5x) = 1-cos(10x)
In that case, it's just as simple as using the Double Angle Formula for cosine. I forgot the parentheses last time so that's how I ended up with a negative answer. Here is a possible correct answer. (There are multiple ways to prove something is true.)
1) 2sin^2(5x) = 1 - cos(10x)
2)            = 1 - cos(2 * 5x)
3)            = 1 - (1 - 2sin^2(5x))
4)            = 1 - 1 + 2sin^2(5x)
5) 2sin^2(5x) = 2sin^2(5x)

This proof basically contains circular logic, going from step 2 to step 3. That step can only be made if the following is true: cos(2*5x)=1-2sin^2(5x). This is basically the same as the original problem, just rearranged a bit. Look at it this way:

1) 2sin^2(5x) = 1 - cos(10x)
2) 2sin^2(5x) + cos(10x) = 1
3) cos(10x) = 1 - 2sin^2(5x)
4) cos(2 * 5x) = 1 - 2sin^2(5x)

Just rearranging things, we can see that this problem is basically asking to prove that the Double Angle Formula is true (EDIT: After thinking on this a bit, maybe its not quite like that. This may just be an exercise of recognizing when and how the Double Angle Identity applies to simple problems. Read on if you care to read the rest of this rant.), so your proof above seems unsatisfactory to me. However, it just may be what the teacher is looking for since the actual proof requires quite a bit of work. But if you assumed that Pythagorean Theorem and Angle Sum Identities are true, then its not too bad.

Angle Sum Identities seem to be tricky to prove, but they state two things:

sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B)
cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)

But we only need the second one. And let's say that A=B, then we get

cos(2A) = cos^2(A) - sin^2(A)

This is one form of a Double Angle Identity. We can use Pythagorean Theorem to substitute cos^2(A) with 1-sin^2(A) and we get an alternate form:

cos(2A) = 1 - sin^2(A) - sin^2(A)
cos(2A) = 1 - 2*sin^2(A)

Now we have arrived at one of the alternate forms of the Double Angle Identity. For this problem in particular we can show that A = 5x:

cos(2*5x) = 1 - 2*sin^2(5x)
cos(10x) = 1 - 2*sin^2(5x)

Which is now just a couple simple algebraic steps away from being the original problem. Not sure if this helped anything. For a proof of the Angle Sum Identity, go here: en.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Angle_sum_identities

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#1 2015-03-20 05:58:43

Math help? c:

#2 2015-03-20 06:13:51

Re: Math help? c:

#3 2015-03-20 07:04:22

Re: Math help? c:

#4 2015-03-20 11:56:14, last edited by Stagecrew (2015-03-20 12:01:26)

Re: Math help? c:

#5 2015-03-20 17:13:20

Re: Math help? c:

#6 2015-03-20 21:57:51

Re: Math help? c:

#7 2015-03-20 23:21:34

Re: Math help? c:

#8 2015-03-20 23:54:49

Re: Math help? c:

#9 2015-03-21 02:50:58

Re: Math help? c:

#10 2015-03-21 06:59:43, last edited by Lionhart (2015-03-21 07:04:23)

Re: Math help? c:

#11 2015-03-21 21:06:09, last edited by rubixguy (2015-03-21 21:15:14)

Re: Math help? c:

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