# Trig Question

## 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.

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
Gold Member
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 Helper

## 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
$$\frac{b}{a}+ \frac{d}{c}$$

$$\frac{sin(x)}{cos(x)}+ \frac{cos(x)}{sin(x)}= \frac{sin^2(x}{sin(x)cos(x)}+ \frac{cos^2(x)}{sin(x)cos(x)}= \frac{sin^2(x)+ cos^2(x)}{sin(x)cos(x)}= \frac{1}{sin(x)cos(x)}$$

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
Gold Member
Just to show what HallsOfIvy was saying:$$\frac{csc(x)}{\frac{sin^{2}(x)+cos^{2}(x)}{sin(x)cos(x)}} = \frac{csc(x)}{\frac{1}{sin(x)cos(x)}} = \frac{\frac{1}{sin(x)}}{\frac{1}{sin(x)cos(x)}} = \frac{sin(x)cos(x)}{sin(x)} = cos(x)$$

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
Gold Member
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.

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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.

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

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

Gotcha, shoulda 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?

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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
Gold Member
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.

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 :tongue2:

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

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
Gold Member
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?

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$$\leq$$x$$\leq$$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)$$\frac{-1-cos(x)}{cos(x)}$$=sec(x)-3 to cos(x)-sec(x)-1=sec(x)-3

MysticDude
Gold Member
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.

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

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

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

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

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.

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[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.

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