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
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