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Hi Phale,

We need to hear the answer to Ray Trent's question.

In the meantime, some identities that may help are:

tan(x) = sin(x)/cos(x), where cos(x) ^= 0

csc(x) = 1/sin(x), where sin(x) ^= 0

We need to hear the answer to Ray Trent's question.

In the meantime, some identities that may help are:

tan(x) = sin(x)/cos(x), where cos(x) ^= 0

csc(x) = 1/sin(x), where sin(x) ^= 0

Cheers!

Dave

Dave

- davidkallman
- Former Expert
**Posts:**675**Joined:**Thu Feb 03, 2005 3:38 pm

Hi Phale,

I suspected that your problem was as such.

I don't know of an elegant solution. The material below is brute force.

Using the equations I posted yesterday, one wants to find x such that:

sin(x)/cos(x)-1/sin(x)=0.

One can calculate approximations to such x by calculating sin(x)/cos(x)-1/sin(x) for values of:

1. x = -3.4 down through -.1 (advancing by -.1) and

2. x = .1 up through 3.4 (advancing by .1)

(sin(x)/cos(x)-1/sin(x) is undefined when x = 0)

You will discover that there are two solutions (x1 and x2), x1>0 and x2<0. It appears that x1=-x2, but I don't how to prove that.

By looking at the calculated values, you'll have calculated x1 and x2 to within .1. You can go through the same process using .01, .001 etc. to compute x1 and x2 to arbitrary precision.

Since the formulas are roughly symmetrical, the above is calculated fairly easily using a spreadsheet.

I'm hesitant to give you the values I've computed (x1 and x2 to within four decimal places) as that is not the style of this bulletin board.

Happy computing. And of course if any one has anything more elegant, please let us know!

I suspected that your problem was as such.

I don't know of an elegant solution. The material below is brute force.

Using the equations I posted yesterday, one wants to find x such that:

sin(x)/cos(x)-1/sin(x)=0.

One can calculate approximations to such x by calculating sin(x)/cos(x)-1/sin(x) for values of:

1. x = -3.4 down through -.1 (advancing by -.1) and

2. x = .1 up through 3.4 (advancing by .1)

(sin(x)/cos(x)-1/sin(x) is undefined when x = 0)

You will discover that there are two solutions (x1 and x2), x1>0 and x2<0. It appears that x1=-x2, but I don't how to prove that.

By looking at the calculated values, you'll have calculated x1 and x2 to within .1. You can go through the same process using .01, .001 etc. to compute x1 and x2 to arbitrary precision.

Since the formulas are roughly symmetrical, the above is calculated fairly easily using a spreadsheet.

I'm hesitant to give you the values I've computed (x1 and x2 to within four decimal places) as that is not the style of this bulletin board.

Happy computing. And of course if any one has anything more elegant, please let us know!

Cheers!

Dave

Dave

- davidkallman
- Former Expert
**Posts:**675**Joined:**Thu Feb 03, 2005 3:38 pm

Hi Phale,

I agree with Dave that there doesn't seem to be a really neat way to do this problem, although there may be something that I am not seeing. Using some trig identities, I was able to wrestle it into a form in which x was alone on one side, although it was still in the other side as well. I put this into my graphing calculator as a function in y. You could try doing something of this sort, and then looking for places on the graph in which x=y, it may help you to zero in on the correct answer slightly faster.

Good luck!

-Emily

I agree with Dave that there doesn't seem to be a really neat way to do this problem, although there may be something that I am not seeing. Using some trig identities, I was able to wrestle it into a form in which x was alone on one side, although it was still in the other side as well. I put this into my graphing calculator as a function in y. You could try doing something of this sort, and then looking for places on the graph in which x=y, it may help you to zero in on the correct answer slightly faster.

Good luck!

-Emily

Reach for the stars and, if you miss, grab the moon!

- EmilyDolson
- Former Expert
**Posts:**27**Joined:**Wed Oct 31, 2007 3:33 pm**Occupation:**Student**Project Question:**Human Impacts on Sea Otter Behavior**Project Due Date:**March 2009**Project Status:**I am conducting my research

davidkallman wrote:Hi Phale,

I suspected that your problem was as such.

I don't know of an elegant solution. The material below is brute force.

Using the equations I posted yesterday, one wants to find x such that:

sin(x)/cos(x)-1/sin(x)=0.

One can calculate approximations to such x by calculating sin(x)/cos(x)-1/sin(x) for values of:

1. x = -3.4 down through -.1 (advancing by -.1) and

2. x = .1 up through 3.4 (advancing by .1)

(sin(x)/cos(x)-1/sin(x) is undefined when x = 0)

You will discover that there are two solutions (x1 and x2), x1>0 and x2<0. It appears that x1=-x2, but I don't how to prove that.

By looking at the calculated values, you'll have calculated x1 and x2 to within .1. You can go through the same process using .01, .001 etc. to compute x1 and x2 to arbitrary precision.

Since the formulas are roughly symmetrical, the above is calculated fairly easily using a spreadsheet.

I'm hesitant to give you the values I've computed (x1 and x2 to within four decimal places) as that is not the style of this bulletin board.

Happy computing. And of course if any one has anything more elegant, please let us know!

Hi, davidkallman

Thanks so much for helping.

I've just asked my mathematical teacher this morning, he solved the problem by using a calculator to graph to function y = tan x and y = csc x. He set the window xmin = -Ï€, xmax = Ï€, ymin = -5, and ymax = 5. Then, he found the intersections. You are correct, there are two solution x = 0.93 and x = -0.93. However, I'd rather to know how to solve it without graphing. Just want to let you know what my teacher did to solve this. Thank you again

Phale

- phale
**Posts:**10**Joined:**Sun Nov 11, 2007 11:05 am

sin(x)/cos(x)-1/sin(x)=0.

If you multiply through by sin(x)cos(x) you get sin(x)**2 - cos(x) = 0.

Using identity sin(x)**2 + cos(x)**2 = 1 and solving for sin(x)**2 and plugging it back in I get:

-cos(x)**2 - cos(x) + 1 = 0

Let z = cos(x) and I get -z**2 - z + 1 = 0

which can be solved by the quadratic equation for z.

You can then use arccos(z) to solve for x.

-Craig

- Craig_Bridge
- Expert
**Posts:**1297**Joined:**Mon Oct 16, 2006 11:47 am

Craig_Bridge wrote:sin(x)/cos(x)-1/sin(x)=0.

If you multiply through by sin(x)cos(x) you get sin(x)**2 - cos(x) = 0.

Using identity sin(x)**2 + cos(x)**2 = 1 and solving for sin(x)**2 and plugging it back in I get:

-cos(x)**2 - cos(x) + 1 = 0

Let z = cos(x) and I get -z**2 - z + 1 = 0

which can be solved by the quadratic equation for z.

You can then use arccos(z) to solve for x.

Hi, Craig

It does make sense! It's not that hard once you get it. You are just smart. Thanks a lot

Phale

- phale
**Posts:**10**Joined:**Sun Nov 11, 2007 11:05 am

It does make sense! It's not that hard once you get it. You are just smart.

Solved problems always look simpler. You really don't have to be smart to figure this one out; however, the experience of going through engineering before graphing calculators required learning how to look for simple transforms to solve test problems quickly which is now mostly a lost art except for math majors with professors who insist on finding a closed form solution.

-Craig

- Craig_Bridge
- Expert
**Posts:**1297**Joined:**Mon Oct 16, 2006 11:47 am

Craig_Bridge wrote:sin(x)/cos(x)-1/sin(x)=0.

If you multiply through by sin(x)cos(x) you get sin(x)**2 - cos(x) = 0.

Using identity sin(x)**2 + cos(x)**2 = 1 and solving for sin(x)**2 and plugging it back in I get:

-cos(x)**2 - cos(x) + 1 = 0

Let z = cos(x) and I get -z**2 - z + 1 = 0

which can be solved by the quadratic equation for z.

You can then use arccos(z) to solve for x.

Craig, thank you. I knew there had to be a closed solution (one with a formula). My brute force solution, while giving an approximation, was the wrong way to go. (one could say the same about using a graphing calculator, which is just a variant of my method.)

This comes from someone with undergraduate and graduate degrees in mathematics.

Cheers!

Dave

Dave

- davidkallman
- Former Expert
**Posts:**675**Joined:**Thu Feb 03, 2005 3:38 pm

Craig_Bridge has the correct solution, beat me to it... dang :).

I'm a little late to the party but....

There is one minor but very important thing to remember. Trig functions are periodic. The approach Craig_Bridge described provides 1 answer (His solution reveals two possible values but only one works and you have to figure out which one).

There is actually another answer within the first per 2*pi cycle of sin() and cos().

So to get the second answer you need to use some trig logic to work out the second solution. (Try plotting the first answer on a unit circle.... you should be able to see the second answer quickly. Or you could use a graphing calculator or even an Excel spreadsheet/graph.)

So once you've figured out the two answers per 2*pi cycle, you must find all of the possible solutions over all cycles such that the answers are within -n and n as defined by your original problem. This will very possibly reveal that there may be 1 valid answer, 2 valid answers or even many more.

I'm a little late to the party but....

There is one minor but very important thing to remember. Trig functions are periodic. The approach Craig_Bridge described provides 1 answer (His solution reveals two possible values but only one works and you have to figure out which one).

There is actually another answer within the first per 2*pi cycle of sin() and cos().

So to get the second answer you need to use some trig logic to work out the second solution. (Try plotting the first answer on a unit circle.... you should be able to see the second answer quickly. Or you could use a graphing calculator or even an Excel spreadsheet/graph.)

So once you've figured out the two answers per 2*pi cycle, you must find all of the possible solutions over all cycles such that the answers are within -n and n as defined by your original problem. This will very possibly reveal that there may be 1 valid answer, 2 valid answers or even many more.

================================

David, Algonquin IL, Geek

David, Algonquin IL, Geek

- davidcastagna
- Former Expert
**Posts:**12**Joined:**Sat Feb 03, 2007 7:43 am

davidcastagna wrote:So once you've figured out the two answers per 2*pi cycle, you must find all of the possible solutions over all cycles such that the answers are within -n and n as defined by your original problem. This will very possibly reveal that there may be 1 valid answer, 2 valid answers or even many more.

Since the original question was find x, where:

tan x = csc x, -Ï€ â‰¤ x â‰¤ Ï€

we can rule out case of even many more.

Cheers!

Dave

Dave

- davidkallman
- Former Expert
**Posts:**675**Joined:**Thu Feb 03, 2005 3:38 pm

Hi! This is my first post, so bear with me

Here's how I'd solve it:

sin x / cos x = 1 / sin x

Cross multiply:

sin^2 x = cos x

Since sin^2 x + cos^2 x = 1,

1 - cos^2 x = cos x

1 - cos^2 x - cos x = 0

Rearrange terms and divide by -1 to reduce the numerator of the cos^2 term to 1:

cos^2 x + cos x - 1 = 0

Use the quadratic formula to solve the solution. Set y = cos x, and you get

y^2 + y - 1 = 0

This has two solutions. You will find that one of them will not work (can you see why)? The other one will work. I can't give you the solution, though.

P.S. If I recall correctly, arccos(y/2) is an integral number of degrees. Do you know what it is?

ACG

Here's how I'd solve it:

sin x / cos x = 1 / sin x

Cross multiply:

sin^2 x = cos x

Since sin^2 x + cos^2 x = 1,

1 - cos^2 x = cos x

1 - cos^2 x - cos x = 0

Rearrange terms and divide by -1 to reduce the numerator of the cos^2 term to 1:

cos^2 x + cos x - 1 = 0

Use the quadratic formula to solve the solution. Set y = cos x, and you get

y^2 + y - 1 = 0

This has two solutions. You will find that one of them will not work (can you see why)? The other one will work. I can't give you the solution, though.

P.S. If I recall correctly, arccos(y/2) is an integral number of degrees. Do you know what it is?

ACG

You know it's cold when the sine is negative.

You know it's hot when the cosine is negative.

You know it's hot when the cosine is negative.

- acgoldis
- Former Expert
**Posts:**6**Joined:**Sat Oct 25, 2008 7:06 pm**Occupation:**Software Engineer**Project Question:**N/A**Project Due Date:**N/A**Project Status:**Not applicable

acgoldis wrote:Use the quadratic formula to solve the solution. Set y = cos x, and you get

y^2 + y - 1 = 0

This has two solutions. You will find that one of them will not work (can you see why)? The other one will work. I can't give you the solution, though.

P.S. If I recall correctly, arccos(y/2) is an integral number of degrees. Do you know what it is?

ACG

Hi AGC,

As you correctly point out, there are two solutions to the equation. One has cos(x) in the [-1,1] interval; so arccos(x) is defined. The other has cos(x) outside this interval. so arccos(x) is not defined.

Two questions:

1. You state "arccos(y/2) is an integral number of degrees". Why? For example, in this case, arccos(y) is non-integral.

2. The last post on this topic was eleven months ago. Why is it being revived?

Cheers!

Dave

Dave

- davidkallman
- Former Expert
**Posts:**675**Joined:**Thu Feb 03, 2005 3:38 pm

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