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

[caharris]

Homework Statement

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

Homework Equations

All of the trigonometric identities. Sin^{2}+Cos^{2}=1; tan^{2}+1=Sec^{2}; 1+Cot^{2}=Csc^{2}; 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) \frac{Csc(x)}{\frac{Sin(x)}{Cos(x)}+\frac{Cos(x)}{Sin(x)}}=Cos(x)
\frac{1}{Sin(x)}*\frac{Cos(x)}{Sin(x)}+\frac{Sin(x)}{Cos(x)}=Cos(x)
\frac{1}{Cos(x)}+\frac{Cos(x)}{Sin(x)}=Cos(x)
*Stuck*

2)Csc(x)=Cos(x)Cot(x)+Tan(x)
Csc(x)=Cos(x)\frac{Cos(x)}{Sin(x)}+\frac{Sin(x)}{Cox(x)}
Csc(x)=Sin(x)+\frac{Cos(x)}{Sin(x)}
*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.

 

[micromass]

The first. The identity

\frac{csc(x)}{cot(x)+tan(x)}=cos(x)

iff

csc(x)=cos(x)(cot(x)+tan(x))

iff

1=sin(x)cos(x)(cot(x)+tan(x))

iff

1=sin(x)cos(x)(\frac{cos(x)}{sin(x)}+\frac{sin(x)}{cos(x)})

I think you can take it from here.




For the second:

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

and I'll leave the rest to you.

 

[MysticDude]

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.

 

[HallsofIvy]

Homework Statement

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

Homework Equations

All of the trigonometric identities. Sin^{2}+Cos^{2}=1; tan^{2}+1=Sec^{2}; 1+Cot^{2}=Csc^{2}; 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) \frac{Csc(x)}{\frac{Sin(x)}{Cos(x)}+\frac{Cos(x)}{Sin(x)}}=Cos(x)
\frac{1}{Sin(x)}*\frac{Cos(x)}{Sin(x)}+\frac{Sin(x)}{Cos(x)}=Cos(x)

This is wrong.
\frac{1}{\frac{a}{b}+ \frac{c}{d}}
is NOT equal to
\frac{b}{a}+ \frac{d}{c}

Instead
\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)}

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

That will make finishing this easy.

\frac{1}{Cos(x)}+\frac{Cos(x)}{Sin(x)}=Cos(x)
*Stuck*

2)Csc(x)=Cos(x)Cot(x)+Tan(x)
Csc(x)=Cos(x)\frac{Cos(x)}{Sin(x)}+\frac{Sin(x)}{Cox(x)}
Csc(x)=Sin(x)+\frac{Cos(x)}{Sin(x)}
*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.

 

[MysticDude]

Just to show what HallsOfIvy was saying:\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)

 

[caharris]

Just to show what HallsOfIvy was saying:\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)

This is what I got, to be honest. Took me four hours, but I got it! Thank you guys so much!

 

[MysticDude]

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 -cos^{4}(x) or the positive.

 

[caharris]

For the second:

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

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.

 

[micromass]

I mean that the left side changes in the right side.

 

[caharris]

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?

 

[caharris]

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.

 

[MysticDude]

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:
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) which is what you get.

 

[caharris]

1-sin²(x) just positive cos²(x).
Anyway look:
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) which is what you get.

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

All of your help was invaluable! Thank you all again.

 

[caharris]

Another quick question: can you divide out a Cos(x) from Cos^{2}(x) - Sin^{2}(x) + 3Cos(x) = 1 and get Cos(x) - \frac{Sin^{2}(x)}{Cos(x)} + 3 = Sec(x) ? Or would it make it Cos(x) - \frac{Sin^{2}(x)}{Cos(x)} + 2Cos(x) = Sec(x) ?

 

[MysticDude]

Another quick question: can you divide out a Cos(x) from Cos²(x) - Sin²(x) + 3Cos(x) = 1 and get Cos(x) - \frac{Sin²(x)}{Cos(x)} + 3 = Sec(x) ? Or would it make it Cos(x) - \frac{Sin²(x)}{Cos(x)} + 2Cos(x) = Sec(x) ?

Yes, the first one is correct. Are you doing a trig proof again?

 

[caharris]

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\leqx\leq360deg" 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)\frac{-1-cos(x)}{cos(x)}=sec(x)-3 to cos(x)-sec(x)-1=sec(x)-3

 

[MysticDude]

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 \frac{cos(x)}{sin^{2}(x)} = \frac{2}{3} 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.

 

[caharris]

Well it's cos2(x), not cos²(x). Double angle formula. (Sorry that I didn't make that clear.) Same with sin as well.

 

[MysticDude]

It seemed as if you intentionally put cos²(x). Oh and I edited my last post just in case.

 

[caharris]

Gosh, I feel so dumb. "Solve cos2(x)+3cos(x)-1=0"
I messed up that whole problem... Give me a few minutes.

 

[MysticDude]

No it's OK since cos(2x) = cos²(x) - sin²(x)!
You didn't make a mistake!

 

[caharris]

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

 

[caharris]

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 \frac{Cos(x)}{Sin^2(x)} = \frac{2}{3} but I don't know how to get that into a single identity to get a degree.

 

[caharris]

[edit:]My teacher did it where she got cos(x)=\frac{1}{2}, but when I tried it I got cos(x)=\frac{-1}{2}.
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)-\frac{-1}{2}
(cos(x)+2)(2cos(x)-1)=0
[STRIKE]cos(x)=-2[/STRIKE] cos(x)=\frac{1}{2}

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

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