Verify "Sin2(x)Cos2(x) - Cos2(x) = 0" Identity

  • Thread starter caharris
  • Start date
  • Tags
    Identity
MysticDudeI'm not sure what you're asking. I was just showing how to simplify the left side of the equation.
  • #1
caharris
13
0

Homework Statement


Verify that [tex]\frac{Csc(x)}{Cot(x)+Tan(x)}[/tex]=Cos(x) is an identity.


Homework Equations


All of the trigonometric identities. Sin[tex]^{2}[/tex]+Cos[tex]^{2}[/tex]=1; tan[tex]^{2}[/tex]+1=Sec[tex]^{2}[/tex]; 1+Cot[tex]^{2}[/tex]=Csc[tex]^{2}[/tex]; etc.


The Attempt at a Solution


I've literally written about five pages worth trying different things, so I'll get you to where I get lost (both ways) that seemed the most promising to me.

1) [tex]\frac{Csc(x)}{\frac{Sin(x)}{Cos(x)}+\frac{Cos(x)}{Sin(x)}}[/tex]=Cos(x)
[tex]\frac{1}{Sin(x)}[/tex]*[tex]\frac{Cos(x)}{Sin(x)}+\frac{Sin(x)}{Cos(x)}[/tex]=Cos(x)
[tex]\frac{1}{Cos(x)}[/tex]+[tex]\frac{Cos(x)}{Sin(x)}[/tex]=Cos(x)
*Stuck*

2)Csc(x)=Cos(x)Cot(x)+Tan(x)
Csc(x)=Cos(x)[tex]\frac{Cos(x)}{Sin(x)}[/tex]+[tex]\frac{Sin(x)}{Cox(x)}[/tex]
Csc(x)=Sin(x)+[tex]\frac{Cos(x)}{Sin(x)}[/tex]
*stuck*
(This didn't type out very well, I'll be happy to post a picture of it worked out if you'd like)

Homework Statement


Simplify Sin2(x) Cos2(x) - Cos2(x)

Homework Equations


Same equations as above.

The Attempt at a Solution


To be honest, I don't even know where to begin. I tried changing all of them into their Pythagorean identities to see if I could end up canceling anything and I don't think I can. I thought about dividing the whole thing by Cos2, but I don't know if I could. Though, when I tried to do that anyways, I still ended up stuck.
I'll post all the trial and error pictures if you'd like.

Any help is greatly appreciated. I don't need any answers, I just would like a suggestion or a nest-step idea. I've been working on these for a few hours and I think I've become numb to see anything new.
 
Physics news on Phys.org
  • #2
The first. The identity

[tex]\frac{csc(x)}{cot(x)+tan(x)}=cos(x)[/tex]

iff

[tex]csc(x)=cos(x)(cot(x)+tan(x))[/tex]

iff

[tex] 1=sin(x)cos(x)(cot(x)+tan(x))[/tex]

iff

[tex] 1=sin(x)cos(x)(\frac{cos(x)}{sin(x)}+\frac{sin(x)}{cos(x)})[/tex]

I think you can take it from here.




For the second:

[tex] \sin^2(x)\cos^2(x)-\cos^2(x)=-\cos^2(1-\sin^2(x))[/tex]

and I'll leave the rest to you.
 
  • #3
For the first problem:
Well have you ever thought about actually adding cot(x) and tan(x)? I did it that way and I was able to get cos(x). I have this tex image that has the answer but let's see if you can get it too.
So the tip was: add the cot(x) and the tan(x).

Micromass is doing it differently, my teacher didn't allow us to use the right side of the equation that much.
 
  • #4
caharris said:

Homework Statement


Verify that [tex]\frac{Csc(x)}{Cot(x)+Tan(x)}[/tex]=Cos(x) is an identity.


Homework Equations


All of the trigonometric identities. Sin[tex]^{2}[/tex]+Cos[tex]^{2}[/tex]=1; tan[tex]^{2}[/tex]+1=Sec[tex]^{2}[/tex]; 1+Cot[tex]^{2}[/tex]=Csc[tex]^{2}[/tex]; etc.


The Attempt at a Solution


I've literally written about five pages worth trying different things, so I'll get you to where I get lost (both ways) that seemed the most promising to me.

1) [tex]\frac{Csc(x)}{\frac{Sin(x)}{Cos(x)}+\frac{Cos(x)}{Sin(x)}}[/tex]=Cos(x)
[tex]\frac{1}{Sin(x)}[/tex]*[tex]\frac{Cos(x)}{Sin(x)}+\frac{Sin(x)}{Cos(x)}[/tex]=Cos(x)
This is wrong.
[tex]\frac{1}{\frac{a}{b}+ \frac{c}{d}}[/tex]
is NOT equal to
[tex]\frac{b}{a}+ \frac{d}{c}[/tex]

Instead
[tex]\frac{sin(x)}{cos(x)}+ \frac{cos(x)}{sin(x)}= \frac{sin^2(x}{sin(x)cos(x)}+ \frac{cos^2(x)}{sin(x)cos(x)}= \frac{sin^2(x)+ cos^2(x)}{sin(x)cos(x)}= \frac{1}{sin(x)cos(x)}[/tex]

So
[tex]\frac{1}{tan(x)+ cot(x)}= \frac{1}{\frac{sin(x)}{cos(x)}+ \frac{cos(x)}{sin(x)}}= sin(x)cos(x)[/tex].

That will make finishing this easy.

[tex]\frac{1}{Cos(x)}[/tex]+[tex]\frac{Cos(x)}{Sin(x)}[/tex]=Cos(x)
*Stuck*

2)Csc(x)=Cos(x)Cot(x)+Tan(x)
Csc(x)=Cos(x)[tex]\frac{Cos(x)}{Sin(x)}[/tex]+[tex]\frac{Sin(x)}{Cox(x)}[/tex]
Csc(x)=Sin(x)+[tex]\frac{Cos(x)}{Sin(x)}[/tex]
*stuck*
(This didn't type out very well, I'll be happy to post a picture of it worked out if you'd like)

Homework Statement


Simplify Sin2(x) Cos2(x) - Cos2(x)

Homework Equations


Same equations as above.

The Attempt at a Solution


To be honest, I don't even know where to begin. I tried changing all of them into their Pythagorean identities to see if I could end up canceling anything and I don't think I can. I thought about dividing the whole thing by Cos2, but I don't know if I could. Though, when I tried to do that anyways, I still ended up stuck.
I'll post all the trial and error pictures if you'd like.

Any help is greatly appreciated. I don't need any answers, I just would like a suggestion or a nest-step idea. I've been working on these for a few hours and I think I've become numb to see anything new.
 
  • #5
Just to show what HallsOfIvy was saying:[tex]\frac{csc(x)}{\frac{sin^{2}(x)+cos^{2}(x)}{sin(x)cos(x)}} = \frac{csc(x)}{\frac{1}{sin(x)cos(x)}} = \frac{\frac{1}{sin(x)}}{\frac{1}{sin(x)cos(x)}} = \frac{sin(x)cos(x)}{sin(x)} = cos(x)[/tex]
 
  • #6
MysticDude said:
Just to show what HallsOfIvy was saying:[tex]\frac{csc(x)}{\frac{sin^{2}(x)+cos^{2}(x)}{sin(x)cos(x)}} = \frac{csc(x)}{\frac{1}{sin(x)cos(x)}} = \frac{\frac{1}{sin(x)}}{\frac{1}{sin(x)cos(x)}} = \frac{sin(x)cos(x)}{sin(x)} = cos(x)[/tex]

This is what I got, to be honest. Took me four hours, but I got it! Thank you guys so much!
 
  • #7
caharris said:
This is what I got, to be honest. Took me four hours, but I got it! Thank you guys so much!

Well, did you also get the second problem? Just from memory I think it was supposed to be [itex]-cos^{4}(x)[/itex] or the positive.
 
Last edited:
  • #8
micromass said:
For the second:

[tex] \sin^2(x)\cos^2(x)-\cos^2(x)=-\cos^2(1-\sin^2(x))[/tex]

and I'll leave the rest to you.

Are you saying that the left changes into the right side of the equation, or the cos2(x) from the original equation turns into the right side of your equation?

@MysticDude
No, I'm still working on it. I'm going to try micromass' suggestion now.
 
  • #9
I mean that the left side changes in the right side.
 
  • #10
micromass said:
I mean that the left side changes in the right side.

Gotcha, should have recognized that. I'm tired, so I probably just missed it. Thank you for your help!

Did you get this by factoring out a Cos2?
 
Last edited:
  • #11
Okay, I'm thinking this is wrong, but here's what I did. I plugged the formula into Mathematica and it gave me Cos2(x)(1-Sin2(x)). I figured out that, via moving the Sin2(x)+Cos2(x)=1 formula around that 1-Sin2(x) is equal to -Cos2(x). I plugged this into the original formula (that Mathematica gave me) and the answer turns into -Cos2(x)2.

I think this would suffice as an answer, but I'm a tad lost on how Mathematica got the formula it did, and if micromass' is any different.
 
  • #12
caharris said:
Okay, I'm thinking this is wrong, but here's what I did. I plugged the formula into Mathematica and it gave me Cos2(x)(1-Sin2(x)). I figured out that, via moving the Sin2(x)+Cos2(x)=1 formula around that 1-Sin2(x) is equal to -Cos2(x). I plugged this into the original formula (that Mathematica gave me) and the answer turns into -Cos2(x)2.

I think this would suffice as an answer, but I'm a tad lost on how Mathematica got the formula it did, and if micromass' is any different.

1-sin²(x) just positive cos²(x).
Anyway look:
[tex]sin^2(x)cos^2(x) - cos^2(x) = -cos^2(x)(1-sin^2(x)) = -cos^2(x)(cos^2(x)) = -cos^4(x)[/tex] which is what you get.
 
  • #13
MysticDude said:
1-sin²(x) just positive cos²(x).
Anyway look:
[tex]sin^2(x)cos^2(x) - cos^2(x) = -cos^2(x)(1-sin^2(x)) = -cos^2(x)(cos^2(x)) = -cos^4(x)[/tex] which is what you get.

Ahh, thank you so much! I wasn't that far off :tongue2:

All of your help was invaluable! Thank you all again.
 
  • #14
Another quick question: can you divide out a Cos(x) from Cos[tex]^{2}[/tex](x) - Sin[tex]^{2}[/tex](x) + 3Cos(x) = 1 and get Cos(x) - [tex]\frac{Sin^{2}(x)}{Cos(x)}[/tex] + 3 = Sec(x) ? Or would it make it Cos(x) - [tex]\frac{Sin^{2}(x)}{Cos(x)}[/tex] + 2Cos(x) = Sec(x) ?
 
  • #15
caharris said:
Another quick question: can you divide out a Cos(x) from Cos2(x) - Sin2(x) + 3Cos(x) = 1 and get Cos(x) - [tex]\frac{Sin2(x)}{Cos(x)}[/tex] + 3 = Sec(x) ? Or would it make it Cos(x) - [tex]\frac{Sin2(x)}{Cos(x)}[/tex] + 2Cos(x) = Sec(x) ?
Yes, the first one is correct. Are you doing a trig proof again?
 
  • #16
Yeah, I'm checking over my homework, and I'm just making sure everything turns out correct.
"Solve cos2(x)+3cos(x)-1=0 for 0deg[tex]\leq[/tex]x[tex]\leq[/tex]360deg" and I got an answer of 315deg. I'm questioning whether that part is correct (which, as you say, it is) and when I somehow went from Cos(x)[tex]\frac{-1-cos(x)}{cos(x)}[/tex]=sec(x)-3 to cos(x)-sec(x)-1=sec(x)-3
 
  • #17
Since you can use the right side of the equation then go ahead and subtract cos²(x) to have -sin²(x) +3cos(x) = 1-cos²(x) which is also the same as -sin²(x) +3cos(x) = sin²(x). In other words: 3cos(x) = 2sin²(x) which is [tex]\frac{cos(x)}{sin^{2}(x)} = \frac{2}{3}[/tex] and you can solve it like that too. I think this way is also correct.

EDIT: I think it would be better to make everything into cos(x). So cos²(x) - (1-cos²(x)) +3cos(x) = 1 which becomes 2cos²(x) + 3cos(x) = 2. Subtracting the 2 gives us a quadratic 2cos²(x) + 3cos(x)-2. From here on you can use the quadratic formula to solve.
 
  • #18
Well it's cos2(x), not cos2(x). Double angle formula. (Sorry that I didn't make that clear.) Same with sin as well.
 
  • #19
It seemed as if you intentionally put cos²(x). Oh and I edited my last post just in case.
 
  • #20
Gosh, I feel so dumb. "Solve cos2(x)+3cos(x)-1=0"
I messed up that whole problem... Give me a few minutes.
 
  • #21
No it's OK since cos(2x) = cos²(x) - sin²(x)!
You didn't make a mistake!
 
  • #22
Wait wait I see what I had done. I turned cos2(x) into cos2-sin2 using the double angle formula. So yeah you're correct.

[edit:]Haha I thought I was about cry because this problem took me forever
 
  • #23
Well, I know my original answer is wrong. I forgot the square when I changed Sin2(x) into 1-Cos2(x). (I had put 1-Cos(x). Ugh.)

[edit:] So I got [tex] \frac{Cos(x)}{Sin^2(x)} = \frac{2}{3}[/tex] but I don't know how to get that into a single identity to get a degree.
 
Last edited:
  • #24
[edit:]My teacher did it where she got cos(x)=[tex] \frac{1}{2}[/tex], but when I tried it I got cos(x)=[tex] \frac{-1}{2}[/tex].
This is what she had (btw, she was kind of out of it, so she may have done something wrong):
2cos2(x)+3cos(x)-2=0
(cos(x)+2)(cos(x)-[tex] \frac{-1}{2}[/tex]
(cos(x)+2)(2cos(x)-1)=0
[STRIKE]cos(x)=-2[/STRIKE] cos(x)=[tex] \frac{1}{2}[/tex]

I, apparently, made an illegal move when I changed [tex] \frac{sin^2(x)}{cos(x)}[/tex] into [tex] \frac{1-(cos(x))(cos(x))}{cos(x)}[/tex] and canceled out one of the cos(x)'s to get three halves equals 1-cos(x), which led me to cos(x)=[tex] \frac{-1}{2}[/tex].

I would assume that she's correct, but she asked me to make sure before I made a final answer.
 
Last edited:

What is the "Sin2(x)Cos2(x) - Cos2(x) = 0" identity?

The "Sin2(x)Cos2(x) - Cos2(x) = 0" identity is a trigonometric identity that states that the product of the sine and cosine of an angle, squared, minus the cosine of the same angle, squared, equals zero.

How is the "Sin2(x)Cos2(x) - Cos2(x) = 0" identity derived?

The identity can be derived using the double angle formula for cosine, which states that cos(2x) = cos^2(x) - sin^2(x). By substituting this into the original equation, we can simplify to get cos^2(x) - cos^2(x) = 0, which proves the identity.

Why is the "Sin2(x)Cos2(x) - Cos2(x) = 0" identity important?

The identity is important because it allows us to simplify and solve trigonometric equations involving the sine and cosine functions. It also has many applications in fields such as physics, engineering, and astronomy.

Can the "Sin2(x)Cos2(x) - Cos2(x) = 0" identity be used to solve all trigonometric equations?

No, the identity can only be used to solve equations where the terms involve the sine and cosine functions, squared. It cannot be used to solve equations with other trigonometric functions, such as tangent or secant.

Are there any exceptions to the "Sin2(x)Cos2(x) - Cos2(x) = 0" identity?

Yes, there are a few exceptions, such as when x is equal to 0, π/2, or π. In these cases, the identity does not hold true and the equation will not equal zero. However, these are the only exceptions and the identity is generally reliable for most values of x.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
4
Views
545
  • Precalculus Mathematics Homework Help
Replies
4
Views
511
  • Precalculus Mathematics Homework Help
Replies
24
Views
2K
  • Precalculus Mathematics Homework Help
Replies
7
Views
979
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
  • Precalculus Mathematics Homework Help
Replies
15
Views
2K
  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
268
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
Back
Top