trigonometric equations

[Stilzkin]

So I suck at math, can anyone help me solve these questions:




tan^2x= 9cos^2x+sin^2x where 0< x < 2 pi, answer in terms of pi

sec^2x-3tanx= -1 here 0< x < 2 pi, answer in terms of pi


Thanks in advance (why isn't there a general math question thread yet?)

[Kaylia]

tan²(x)= 9*cos²(x)+sin²(x)

Just to be sure, this is what you're asking? I can't see any simple method to solve it.

[Stilzkin]

tan²(x)= 9*cos²(x)+sin²(x)

Just to be sure, this is what you're asking? I can't see any simple method to solve it.

yes that's what I meant

[Kaylia]

Might be the sleep pill, but I can't find the identities I need to solve this. It would be easy to solve it numerically and convert the number to pi after, but it's not what you're looking for.

http://en.wikipedia.org/wiki/List_of...ric_identities
sec²(x)-3tan(x)= -1
1/cos²(x)-3sin(x)/cos(x)=-1
3sin(x)cos(x)= cos²(x) +1
3sin(2x)= 2cos²(x) +2
1+cos(2x)-3sin(2x)= -1
cos(2x)-6sin(2x)=-2

,,,dunno, probably screwed up somewhere.

[Woozie]

I don't know any general methods to solving problems like this. When I see problems like this, I just play around with it until something works. So I can't give you advice. The best I can do is solve it for you, and hope that what I did helps you to solve other problems in the future.

http://latex.codecogs.com/gif.latex?...^2(x)+sin^2(x)

Converting everything into sin's, I get:

http://latex.codecogs.com/gif.latex?...2(x))+sin^2(x)

http://latex.codecogs.com/gif.latex?...2(x))+sin^2(x)

http://latex.codecogs.com/gif.latex?...%209-8sin^2(x)

http://latex.codecogs.com/gif.latex?...))(1-sin^2(x))

http://latex.codecogs.com/gif.latex?...2(x)+8sin^4(x)

http://latex.codecogs.com/gif.latex?...2(x)+8sin^4(x)

http://latex.codecogs.com/gif.latex?...2(x)+8sin^4(x)

Define: http://latex.codecogs.com/gif.latex?k\equiv%20sin^2(x), and put this in our equation

http://latex.codecogs.com/gif.latex?8k^2-18k+9=0

so

http://latex.codecogs.com/gif.latex?...0k^2%20=%203/2

Since 3/2 is greater than 1, we can ignore the second solution (because k = sin^2, thus, k can never be greater than 1 because sin^2 is never greater than 1). So we're left with

http://latex.codecogs.com/gif.latex?sin^2(x)%20=%203/4

http://latex.codecogs.com/gif.latex?...rac{\sqrt3}{2}

http://latex.codecogs.com/gif.latex?...m\frac{\pi}{3}

You can (and should) check that both of these solutions are indeed correct.

I think the second one can be solved the same way. Just be sure to go back and check your final answers when you're done. You may end up with extra solutions that don't work. This probably isn't the quickest way to solve these problems, but I really suck at trig identities, so this is the best I can do.

Edit: Actually, the second one is pretty easy. Use 1+tan^2 = sec^2. Then convert it to a polynomial like I did in the above. Except you wont have k = sin^2. You'll have something else (I'll leave it to you to figure it out).

One tip I have for using trig identities is to remember the sin^2+cos^2 = 1 relationship. Not only will this be incredibly useful in any math you take in the future, but also, the tan/sec and cot/csc identities can be derived from it. If you want the tan/sec identity, you just divide both sides by cos^2, and if you want the cot/csc identity, you divide both sides by sin^2. This way you don't have to try to remember all three identities.

Your final answer should be (don't click the spoiler until you work the problem out yourself)

Spoiler: show

http://latex.codecogs.com/gif.latex?\pi/4
http://latex.codecogs.com/gif.latex?5\pi/4
and
http://latex.codecogs.com/gif.latex?tan^{-1}(2)

Again, you should check each answer to verify.

[Max™]

http://www.wolframalpha.com/input/?i...n%C2%B2%28x%29

It is often helpful to plunk things into wolframalpha and see it displayed differently/simplified.

[Eliseos]

http://www.wolframalpha.com/input/?i...n%C2%B2%28x%29

It is often helpful to plunk things into wolframalpha and see it displayed differently/simplified.

Problem with that is you can't use it on an exam, and you don't really learn the identities (or the major ones anyway) and the tricks to solve some of the problems. You just get used to what wolfram tells you. It can help for homework problems though.

[willriker]

I dont know many people who still remember thier trigonometric identities anyhow. Honestly, they have books of trigonometric identities, heck they are even up on the wikipedia. Most schools, i've been to even give you a sheet with trigonometric identities on it for use during exams.

I see no harm using other sources for doing this.

[Woozie]

Well, if your source is the internet, you'll be screwed when exam time comes around. The website max posted rewrites the identity in a lot of ways which could be useful, but wouldn't be available for the test even if he had a table of identities or a trig book in front of him.

[Ramor]

i was always taught to be able to derive the trig identities so i'd never need a sheet, i found out that as much as i hated doing that in high school it helped out a loooot once i got to some of my later calculus courses

[Max™]

Well, I meant that it can be useful to see how the identities were re-written in order to understand them.

Simply solving them teaches you one way to find an answer, learning how to rewrite them, how to take it apart and put it back together, teaches you how to find any answer.

Play around with stuff like that wolframalpha site til you can see how it converted from your input to the various outputs, you'll be much better off later on, as Ramor and others said.

[Eliseos]

Play around with stuff like that wolframalpha site til you can see how it converted from your input to the various outputs, you'll be much better off later on, as Ramor and others said.

Somehow I don't think that is what he was saying, but ok.

[Stilzkin]

Since 3/2 is greater than 1, we can ignore the second solution (because k = sin^2, thus, k can never be greater than 1 because sin^2 is never greater than 1). So we're left with

http://latex.codecogs.com/gif.latex?sin^2(x)%20=%203/4

http://latex.codecogs.com/gif.latex?...rac{\sqrt3}{2}

http://latex.codecogs.com/gif.latex?...m\frac{\pi}{3}

You can (and should) check that both of these solutions are indeed correct.

I think the second one can be solved the same way. Just be sure to go back and check your final answers when you're done. You may end up with extra solutions that don't work. This probably isn't the quickest way to solve these problems, but I really suck at trig identities, so this is the best I can do.

Pi/3 isn't the answer though is it? There is more than one root (is that the right word?) between 0 and 2pi. Even when I manage to solve the equation I'm not sure what to do with the answer.
Is it:
0+pi/3, pi+ pi/3?

[Max™]

Somehow I don't think that is what he was saying, but ok.

i was always taught to be able to derive the trig identities so i'd never need a sheet, i found out that as much as i hated doing that in high school it helped out a loooot once i got to some of my later calculus courses

If you can't see how various sites go from your input, to their output, learning how they did it will do far more for you than just solving each problem one at a time.


Solve a trig problem for a man, he'll get a correct answer on his test, teach a man to solve trig problems, he'll get all the answers correct.

I tried to work in the "build a man a fire, he's warm for a day, set a man on fire, he's warm for the rest of his life" joke, but it didn't work out.

[Eliseos]

Now you're saying the exact opposite of what you originally said. When you put in a trig function into wolfram, they don't show you the steps or the identities used to get to the end result. They only show the end result of an alternate form. There is no way to just get used to "how they go from input to output" if you don't know the trig identities used along the way. Hell, even with complicated problems it'd be impossible for anyone to see the end results without working through them.

[Max™]

Play around with stuff like that wolframalpha site til you can see how it converted from your input to the various outputs

I never said he should only use that site, by Thor man, I simply didn't specify the intent of my statement clearly enough, as I mistakenly assumed it was as obvious for all as it was for me.

All I meant to say was that it is good to have an example of how the equation is worked out so you can "touch base" and make sure you understand how they got there.

Ideally wolfram would spit out all the various steps like woozie did, but being able to at least make sure you're on the right track after trying to work it out yourself is useful, and much more useful than simply having the answer plopped in your lap.


Then, if/when you get stuck, it's good to be able to point to particular steps so those with a better grasp of the subject can determine how much you know, and explain things easier.

[Woozie]

Pi/3 isn't the answer though is it? There is more than one root (is that the right word?) between 0 and 2pi. Even when I manage to solve the equation I'm not sure what to do with the answer.
Is it:
0+pi/3, pi+ pi/3?

My bad, I didn't even realize they wanted answers from 0 to 2pi. Well, if -pi/3 is a solution, then 2pi+(-pi/3) is certainly a solution since sine has a period 2pi. 2pi+(-pi/3) = 5pi/3, so the answers would be pi/3 and 5pi/3