Pre-Calculus Proof

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-Knight-

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I was not sure if I should post this in the trig section or in this section. I figured I would post it here in the calculus section because it is a problem I have run into in my pre calc class.

I'm having major trouble on a proof where I have to prove sin(2x) = 2sin^2(x+(?/4))-1

So far I have spent a couple hours on the problem just working it through different theorems my teacher gave me.

Here are the theorems: (@= theta , ß= beta)

Sum or Difference of two angles:
sin (@+ß) = (sin(@)*cos(ß)) + (cos(@)*sin(ß))
sin (@-ß) = ((sin(@)*cos(ß))-(cos(@)*sin(ß))

Double Angle:

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

Half Angle:

sin^2(@) = 1/2(1-cos(2@))
cos^2(@)= 1/2(1+cos(2@))

Pythagorean Identity:
sin^2(@)+cos^2(@)=1

As I have said I have spent alot of time on this proof, I think i came close to solving it so I will post down the work I have done that came close to solving it.

sin(2x) = 2sin^2(x+(?/4))-1

*Double Angle Theroem on the left. Sum of two angles on the right. I did this to spread out the proof a little bit.

2sin(x) * cos(x) = 2((sin(x)*cos(?/4))+(cos(x)*sin(?/4))^2 -1( 1 not included in exponent)

Next two steps are lengthly. So i did not show the work.
* The next thing I did was I divided everything by cos(x)
This eliminated a cos(x) on the left and right side.
*It also leaves everything else over cos(x). because everything is all over cos(x) we can multiply by cos(x) and have the following:

2sin(x) = 2((sin(x)*cos(?/4))+sin(?/4))^2 -1( 1 not included in exponent)

*I then square everything and FOIL out the 2

2sin(x) = (2sin^2(x)*2cos^2(?/4)) + 2sin^2(?/4) -1
| |
| |______ 2sin^2(?/4) sin(?/4) = root 2 over 2 which is then sqaured and = .5 which is then multiplied by 2 so it =1
|
|______________ 2cos^2(?/4) cos(?/4) = root 2 over 2 which is then squared and = .5 which is then multiplied by 2 so it =1

therefore:

2sin(x) = (2sin^2(x) * 1) +1 -1

you multiply first.

2sin(x) = 2sin^2(x)+1 -1

The 1's cancel

2sin(x) = 2sin^2(x)

and bam. thats what I get :(

I think I might have done something wrong in the very beginning. My teacher said I was making it over complicated by touching the left side. He also said it can be solved in 4-6 steps. Anyone able to help me out? It would be great if someone could show me what I did wrong and possibly help me solve it :) I have spent so long on it and it disappointed me so much when I found out I had an extra square in the answer...

Also this is only one of the many things I tried. It may not be the one that is the closest work wise either. I just know that the answer in the end looked most like the answer I need. Anyways Hope I was thorough enough so that someone can help me without too much trouble :)

Thanks,
Knight

*Edit* I was reading through some of the other forums to see if their was anyone out there I could help out and I see that pre-calc is included under the "Intermediate/advanced Algebra" forum. Feel free to move my post :)
 
If I understand correctly, you had

\(\displaystyle 2\sin x \cos x = 2(\sin x \cos \frac{\pi}{4}+\cos x \sin \frac{\pi}{4})^2 -1\)

Then somehow divided by cos(x), which gave

\(\displaystyle 2\sin x = 2(\frac{\sin x \cos \frac{\pi}{4}}{\cos x}+\sin \frac{\pi}{4})^2 -\frac{1}{\cos x}\)

Then you decided that you can drop the cos(x)'s on the bottom line, to give

\(\displaystyle 2\sin x = 2(\sin x \cos \frac{\pi}{4}+\sin \frac{\pi}{4})^2 -1\)

Neither of these steps is correct.

I'd suggest, instead, remembering the values of cos pi/4 and sin pi/4, then expanding the squared bracket...
 
I know the values for cos pi/4 and sin pi/4 (root2 over 2)

Now by expanding the squared bracket you mean just squaring everything inside?

so

Apply the ^2 to all the terms on the right side

2sin(x)cos(x) = 2(sin^2(x)cos^2(?/4) + cos^2(x)sin^2(?/4))-1

cos and sin of pi/4 = root2 over 2

root2 over 2 sqaured = .5

2sin(x)cos(x) = 2(sin^2(x)(.5)+ cos^2(x)(.5))-1

rewrite

2sin(x)cos(x) = 2((1/2)sin^2(x)+ (1/2)cos^2(x))-1

FOIL the 2

2sin(x)cos(x) = (sin^2(x)+ cos^2(x))-1


I still have a feeling im doing something wrong. Could you please show me an example of your suggestion? It would help me understand better
Also. Do you see it as easier if i just started over without touching the left side? if so, could you point me in the right direction from the very beginning of the proof?

Thanks for your reply at such an early hour :),
Knight

P.S. you were understanding correctly on what I had. (no matter how hard I tried it still looked funny on the computer )
 
This is not that bad if we use some identities. I think you may be making it a little too difficult.

Prove \(\displaystyle sin(2x)=2sin^{2}(x+\frac{\pi}{4})-1\)

Look at the \(\displaystyle 2sin^{2}(x+\frac{\pi}{4})=(cos(x)+sin(x))^{2}\)

Now, expand \(\displaystyle (cos(x)+sin(x))^{2}=2sin(x)cos(x)+1\)

Put it all together and see what the right side is?. The 1's cancel and we have \(\displaystyle 2sin(x)cos(x)=sin(2x)\), which is a well-known identity.
 
I think I'm drawing closer to fully understanding this proof.

I 100% understand the result you got. It is a double angle identity, so that means it is easily provable at that point

But I'm still a little dumbfounded on how you got their.

Ok so we start with:

1.) sin(2x) = 2sin^2(x + ?/4) -1

2sin^2(x + ?/4) = (cos(x) + sin(x))^2

2.) therefore: sin(2x) = (cos(x) + sin(x))^2 -1(Not apart of the exponent)

And then (cos(x) + sin(x))^2 = 2sin(x)cos(x)+1

3.) therefore: sin(2x) = 2sin(x)cos(x)+1-1

1's cancel

4.) sin(2x) = 2sin(x)cos(x) which is a double angle identity

I am pretty sure this is what you meant. Am I understanding correctly? I think what im stuck on is how you go from step 1 to 2. and from 2 to 3.

Where did the coefficient of 2 go in step 1-2?
I thought (cos(x) + sin(x))^2 = 1? (steps 2-3)

How do you know those equal each other? (I 100% believe you that they do. I just want to see the work behind it, so i can better understand.)
I looked at this identity sheet and didn't see anything to do with "2sin^2(x + ?/4) = (cos(x) + sin(x))^2" and "(cos(x) + sin(x))^2 = 2sin(x)cos(x)+1"

could you start with the original equation and then rewrite that equation every time you switch it around, So I can analyze every little thing you did?

Thank you for responding galactus, your help is greatly appreciated and I am making progress in understanding this proof.
 
\(\displaystyle sin(2x) = 2sin^{2}(x+\frac{\pi}{4})-1\)

\(\displaystyle sin(2x) = 2[sin(x)cos(\frac{\pi}{4})+cos(x)sin(\frac{\pi}{4})]^{2}-1\)

\(\displaystyle sin(2x) = 2[\frac{sin(x)+cos(x)}{\sqrt2}]^{2}-1\)

\(\displaystyle sin(2x) = 2[\frac{sin^{2}(x)+2sin(x)cos(x)+cos^{2}(x)}{2}]-1\)

\(\displaystyle sin(2x) = 2sin(x)cos(x)+1-1\)

\(\displaystyle sin(2x) = sin(2x), \ Q.E.D.\)
 
Big glenn, that is very helpful. I understand everything except for how you go from step 2 to step 3. Im going to take some time to analyze it and see if I can figure out what you did on my own. That was VERY helpful and once i totally understand step 2 to step 3 I will be where I want to be with this proof. Thanks a bunch!

If i can wrap my head around it before you read this then I will make and edit! thanks again !!!

edit#1 So i just thought about it for a couple minutes. cos(pi/4) = root2 over 2 = sin (pi/4)
so it turns into:

sin(2x) = 2[sin(x)(sqrt(2)/2) + cos(x)(sqrt(2)/2)]^2

multiplying by root2 over 2 is the same as dividing by the reciprocal? 2 over root2? Something of that nature? is how your getting root2 on the bottom?

edit #2

sin(2x) = 2[(sin(x)/(2(sqrt(2))) + (cos(x)/(2/(sqrt(2)))]^2

that doesnt look that great. but basically it is saying sin(2x) = 2[ (sin(x) divided by (2/root2)) + (cos(x) divided by (2/root2))]^2

and 2/root2 = root2.

and since their both over root2 you can combine them over the same denominator.


WOOT! I figured out how you did step 2-3



Knight
 
I wanted to ask one more question that isnt math oriented. :)

You guys show your work with images. How do you do this? I want to know so that when I help others my work looks better.

Thanks again to all those who helped me with the proof. I 100% understand it now!!

Knight
 
\(\displaystyle sin(\frac{\pi}{4}) = \frac{1}{\sqrt2} \ and \ cos(\frac{\pi}{4} )= \frac{1}{\sqrt2}\)

Knight, to get these forms used Latex (I'm still learning it), check my quote.
 
I never knew that you could think of the sin and cos of a 45 degree angle as 1 over root2. I got root 2 over 2 drilled into me when i hit trig :D

Now i will store that little piece of info away and im sure it will help me in the future.

It must be like html headers, except it is for math?

Should i just google latex? to learn the code?
 
\(\displaystyle \frac{1}{\sqrt2} = \frac{1}{\sqrt2}(\frac{\sqrt2}{\sqrt2}) = \frac{\sqrt2}{2}\)

This is a throw back before calculators, as it was easier to evaluate irrational numbers if one first rationalize the denominator, as which one would you rather do manually: 1.414... divided into 1 or 2 divided into 1.414....?

We still keep it, but it really isn't necessary with the advent of calculators.
 
-Knight- said:
I was not sure if I should post this in the trig section or in this section. I figured I would post it here in the calculus section because it is a problem I have run into in my pre calc class.
I'm having major trouble on a proof where I have to prove sin(2x) = 2sin^2(x+(?/4))-1
Click to expand...

You know

cos(2a) = 1 - 2sin[sup:2z12xf0p]2[/sup:2z12xf0p]a

also

cos(pi/2 + a) = - sin(a)

Then

2sin[sup:2z12xf0p]2[/sup:2z12xf0p](x+(?/4))-1

= - cos[2((x+(?/4))]

= - cos[?/2 + 2x]

= -[sin(2x)]

= sin(2x)
 
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