# Homework Help: Pythagorean Identities and Equations

1. Feb 13, 2010

### Xhanger

1. The problem statement, all variables and given/known data
12cot2(2x)=4

3cos2(3x)=2sin(3x)

cos(3x)=sec(3x)+2tan2(3x)
(6 exact answers) hint(get all cosines) <- teacher wrote that

And then I have to prove by working on one side:
sin3(x) / 1-cosx = sinx*cosx+sinx
and
(sinx / cosx-1) + (sinx / cosx+1) = 2cosx / -sinx

2. Relevant equations
sin^2x + cos2x = 1 (and the other forms)
1+cot2x = csc2x
1+tan2x = sec2x
I can only use these forumulas, however you can rearrange the forumlas

3. The attempt at a solution
There were other problems of the prove by working on one side that I was able to figure out, so just a few beginning steps would do.
For:
12cot2(2x)=4
This is my work, but its not right:
cot2(2x)=1/3
cot2(x)=1/3 (I changed 2x to x so that it would be easier, later I divide my answers by 2)
tan2x=3
tanx=+or-(squareroot(3))
tanx= (sin(sqroot(3)/2)) / cos(1/2) (this is where I divide by 2 because I made 2x -> x above)
So i get answers of +or- pi/3, 2pi/3, 5pi/3 which obviously isn't 8 answers...

Then for:
3cos2(3x)=2sin(3x)
I know how to start it by changing 3cos to 3 - 3sin using one of the pythag theories
but after that I am pretty much clueless

However on this one:
cos(3x)=sec(3x)+2tan2(3x)
(6 exact answers) hint(get all cosines) <- teacher wrote that (however someone told me that sec may be easier)

I am clueless on how to start it, which is the main problem. For most of these problems all I need is alittle help on how to start them.

Then on the ones that I need to prove by working on one side only:
sin3(x) / 1-cosx = sinx*cosx+sinx
and
(sinx / cosx-1) + (sinx / cosx+1) = 2cosx / -sinx

I only need a little help on how to start them or on how to go about solving them. As I said above I already solved a few of these on the paper, only I can't seem to grasp on how to start these specific ones.

Any help at all is appreciated and if anything is confusing please post and I will answers it ASAP.
xhanger

2. Feb 13, 2010

### Mentallic

You know you're allowed to make separate threads for each question
It does make things much simpler too as there's less clutter.

1)
$$12cot^2(2x)=4$$

I'm assuming the restriction on x is $0\leq x < 2\pi$ and this seems to be the case since the question is asking for 8 solutions. This means that $0\leq 2x < 4\pi$ if we multiply everything by 2.

When you make the substitution to get rid of the 2x, make it another variable such as y. Don't use the same variable x.

$$tany=\pm \sqrt{3}$$ (y=2x) after this step I'm unsure what you have done.

For $tany=\sqrt{3}$ we have the answers $2x=\pi/3, 4\pi/3, 7\pi/3, 10\pi/3$ for $0\leq 2x < 4\pi$ and now we can divide through by 2 and you'll notice that all these solutions now fall into the range $0\leq x < 2\pi$

Doing the same for $tany=-\sqrt{3}$ you should get your 8 solutions.

2)
$$3cos^2(3x)=2sin(3x)$$

once you substitute $cos^2(3x)=1-sin^2(3x)$ you'll have a quadratic in $sin(3x)$. Solve it like you would any other quadratic.
To make things clearer, try substitute $y=sin(3x)$ and then substitute back later.

3)
$$cos(3x)=sec(3x)+2tan^2(3x)$$
Have you even tried to take advantage of your teacher's hint? :tongue:

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

$$tan^2(3x)=sec^2(3x)-1$$

Multiply through by cos(3x) to get rid of the sec(3x) etc.

4)
$$\frac{sin^3x}{1-cosx}=sinxcosx+sinx$$

The idea with these proofs is to notice that the LHS has a fraction, while the RHS doesn't. This means you need to find a way to cancel out the denominator $1-cosx$.

Notice that $sin^3x=sinx(sin^2x)$ and $sin^2x=1-cos^2x$.
Now, factorize by using the difference of two squares, and then things should simplify nicely.

5)
$$\frac{sinx}{cosx-1}+\frac{sinx}{cosx-1}=\frac{2cosx}{-sinx}$$

Work with the LHS since it's always easier to go from something more complicated to something simpler.

Notice how the LHS has 2 fractions while the RHS only has 1. This idea should be clear enough in itself! :tongue:

3. Feb 13, 2010

### Xhanger

Very nice response! And quick. Well actually a lot quicker than I thought . Reading though it, it seems that you have cleared up nearly all my questions, but right now its too late for me to be looking into it too deeply. Tomorrow I may be able to look at it in greater detail, but it might have to be pushed back till Sunday till I check it over in detail due to prior plans with friends. Ill post back once I have looked it over in greater detail.
Others are still welcome to comment and give other hints

thanks once again
xhanger

4. Feb 14, 2010

### Xhanger

1. I got all the answers except I cant figure out how you got 7pi/3. Could you explain how you got that one, because I got all the others.

2. I solved the quadratic and got (y+3) and (y-1) which are sin(3x)=-3 and sin(3x)=1. I can only get 2 answers out of that. Could you explain how to get the 6 answers.

3. I don't think I did this right. I got cos2(3x)=2. Could you confirm?

4. Solved it

5. I think you might of wrote this one down wrong in you post. The denominators are cosx-1 and cosx+1. Should I get a common denominator first and work from there? Because I am totally lost on this problem.

5. Feb 15, 2010

### Mentallic

I'll rewrite the solutions in a way that will make more sense.

$$\pi/3, 4\pi/3, 2\pi+\pi/3, 2\pi+4\pi/3$$

Noo! You're solving $y^2+2y-3=(y+3)(y-1)$ when you're meant to be solving $3y^2+2y-3$.
$$3cos^2x=3(1-sin^2x)=3-3sin^2x$$
And think again about why there would be 6 answers. Take a look at what I said earlier in response to post #1

You're very close. Show me step by step what you did and I'll let you know where you went wrong.

Yeah sorry that was a typo. Of course! That's exactly what I was implying haha :tongue:

It should be obvious that you need to find the common denominator because the LHS has two fractions while the RHS only has one. After this just simplify and use one of your trig identities and things should fall in place.

6. Feb 15, 2010

### Xhanger

I got
cos(3x)=(1/cos(3x)) + 2sec2(3x)-2
then I multiplied everything by cos(3x)/1 and got
cos2(3x)=1 + 2sec(3x) - 2cos(3x)
I looked over my paper and 3 obviously isn't right, but I can't figure out what happens with the 2sec2(3x)-1

7. Feb 15, 2010

### Mentallic

Sorry, you weren't close the first time, and neither was I since I made some really stupid error :yuck:

Yes from what you got:

$$cos^2(3x)=1+2sec(3x)-2cos(3x)$$

Notice you still have a sec(3x) to deal with, so you need to multiply through by cos(3x) again. You'll have a cubic, but it has rational roots so it should be easy to solve with the rational root theorem

8. Feb 15, 2010

### Xhanger

Thanks for answering my questions! I finally get it!
Sorry if I bothered you but I wasn't at class for a few days and I got behind.

thanks,
Codester93