Need Help with a Trig Problem

[94FiveOhCOBRA]

I have this trig problem which i'm sure some of you math geniuses will think is easy but i can't find any example of the identities to simplify this problem

Simplify 1 / sin t cos t - 1 / tan t using identities.

1 1
----------- - -------
sin t cos t tan t​

 

[DKS2814V]

Nice, I'm on it... Haven't done one of these in a while...

 

[Coiled03]

I'm coming up with 1- cos^2(t)

Explanation:
1) 1/sin(t)cos(t) - 1/tan(t)
2) 1/sin(t)cost(t) - 1/(sin(t)/cos(t)) tan = sin/cos
3) 1/sin(t)cost(t) - cos(t)/sin(t) invert and multiply
4) 1- cos^2(t) multiply through by sin(t)cos(t) and simplify

Dunno if I messed that up, or if it's what you're looking for, but that's what I came up with.

 

[DKS2814V]

Ditto.....

I just multiplied the left side by tan (t)..... Which is sin (t) / cos (t).....

Which in the end all equals sin^2 (t) because of the identity cos^2 + sin^2 = 1 which can boil down to 1-cos^2 = sin^2


Some of these I had back in school they gave me the answers, just needed to show the work on how to get there. Helps to figure out how simple you want it....

 

[Coiled03]

Ditto.....

I just multiplied the left side by tan (t)..... Which is sin (t) / cos (t).....

Which in the end all equals sin^2 (t) because of the identity cos^2 + sin^2 = 1


Some of these I had back in school they gave me the answers, just needed to show the work on how to get there. Helps to figure out how simple you want it....

DOH! I forgot the last step. :bash:

You're right.....it's sin^2(t)

 

[YELOSNK]

Oh man, please don't remind me of math. I finished calc 2 senior year in HS and I never want to see another math problem that involves more than + - / = ever again. :bash:

 

[BmoseleyINC]

Oh man, please don't remind me of math. I finished calc 2 senior year in HS and I never want to see another math problem that involves more than + - / = ever again. :bash:

haha, I feel you bro. Math is the one subject I suck balls at, probly becasue it's so boring and I never paid attention in class. I was too busy talking to girls and counting how many dots were in the ceiling panels.

I kick ass at Geometry, but calc/trig..forget it.

Anything else, I got you covered and then some. Math? Sorry dude, ask someone else. :bash: :lol:

 

[UCBeau]

I'm coming up with 1- cos^2(t)

Explanation:
1) 1/sin(t)cos(t) - 1/tan(t)
2) 1/sin(t)cost(t) - 1/(sin(t)/cos(t)) tan = sin/cos
3) 1/sin(t)cost(t) - cos(t)/sin(t) invert and multiply
4) 1- cos^2(t) multiply through by sin(t)cos(t) and simplify

Dunno if I messed that up, or if it's what you're looking for, but that's what I came up with.

thats what i got too

damn its been a long time since i did that shit.

 

[94FiveOhCOBRA]

thanks guys

 

[sunburned]

Hey trig, thats fun...

Good solving though.

 

[OZ Dude]

I actually used trig for the first time in about 15 years a few months ago when I needed to figure out the pitch of my roof to calculate the amount of metal I would need to re-roof. I was like :banana: Finally a use for it!! :lol1:

Glad you got the answer you needed :beer:

 

[BittenInBama]

Yeah I worked for a year at a metal building manufacturer; used trig everyday to figure rafter lengths, roof panel lengths, etc. Nice to know it really can be put to a good use every now and then.

 

[killspray]

I think you are all wrong!!!! :nono:

Yes, the 1-Cos^2(t) is correct but you forgot the denominator.

-----------------------------------------------------------------

[1/Sint*Cost] - [1/Tant]

[1/Sint*Cost] - [1/(Sint/Cost)]

[1/Sint*Cost] - [Cost/Sint] **Then multiply [Cost/Sint] by Cost/Cost** (For a common denominator)

[1/Sint*Cost] - [Cos^2(t)/Sint*Cost] **Next, subtract numerators and KEEP denominator**

[1-Cos^2(t)] / [Sint*Cost] **Sin^2(t) + Cos^2(t) = 1** ==> 1-Cos^2(t) = Sin^2(t)

[Sin^2(t)] / [Sint*Cost]

[Sint/Cost]


Answer: Tant


:beer:

 

[killspray]

Are you in college? What's your major??

 

[Coiled03]

I think you are all wrong!!!! :nono:

Yes, the 1-Cos^2(t) is correct but you forgot the denominator.

-----------------------------------------------------------------

[1/Sint*Cost] - [1/Tant]

[1/Sint*Cost] - [1/(Sint/Cost)]

[1/Sint*Cost] - [Cost/Sint] **Then multiply by Cost/Cost**

[1/Sint*Cost] - [Cos^2(t)/Sint*Cost] **Next, subtract numerators and KEEP denominator**

[1-Cos^2(t)] / [Sint*Cost] **Sin^2(t) + Cos^2(t) = 1** ==> 1-Cos^2(t) = Sin^2(t)

[Sin^2(t)] / [Sint*Cost]

[Sint/Cost]


Answer: Tant


:beer:

Ummm.....wrong.

First of all, why would you multiply by cos(t)/cos(t) (which is the same as multiplying by 1...what good does that do?) when multiplying through by sin(t)*cos(t) does the job? Second, you did your multiplication wrong. {[1/Sint*Cost] - [Cost/Sint]} * (cost/cost) = [1/Sint*Cos^2t] - [Cos^2t/Cost*Sint] which simplifies to the exact same thing you started with since all you did is multiply by 1.

sin^2(t) is the right answer

 

[sunburned]

Haha, Killspray is definitely right. The answer is in fact tan(t).

----------------------------------------------------------------

1/(sint*cost)-1/tant

1/(sint*cost)-1/(sint/cost)

1/(sint*cost)-(cost/sint)

Then find a common denominator, which is sint*cost so...

1/(sint*cost)-cos^2(t)/(sint*cost)

(1-cos^2(t))/(sint*cost) : 1-cos^2(t)=sin^2(t)

sin^2(t)/(sint*cost) : cancel out the sint on top and bottom

sint/cost : ** tant **

And my major is mechanical engineering, been doin trig for like 5 years straight. Oh yeah, and I just did the problem on my TI-89, the answer is definitely tan(t). Doesn't anyone use calculators?

 

[killspray]

Ummm.....wrong.

First of all, why would you multiply by cos(t)/cos(t) (which is the same as multiplying by 1...what good does that do?) when multiplying through by sin(t)*cos(t) does the job? Second, you did your multiplication wrong. {[1/Sint*Cost] - [Cost/Sint]} * (cost/cost) = [1/Sint*Cos^2t] - [Cos^2t/Cost*Sint] which simplifies to the exact same thing you started with since all you did is multiply by 1.

sin^2(t) is the right answer

First of all, as I learned in 4th grade, when adding and subtracting fractions, the first thing you have to do is find a common denominator :burn: So I used [Sint*Cost] for my common denominator.

Second, I did not multiply wrong. . . . you did. I only multiplied the second part by [Cost/Cost]. And you are correct. Yes, it does simplify back to the same thing but I did that so I could have a COMMON DENOMINATOR :kaboom:

If you think I am wrong, do a test. Get out your calculator. And for the original equation, substitute in a number for "t". Write that answer down on a piece of paper. Then, plug in that same number, "t", into everyone elses equation, including mine, and tell me who's numbers match the original one.

How can you tell me I am wrong without even checking for yourself if the answer is right?? oke:


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Thank you "Sunburned".

I too am a Mechanical Engineering major. This stuff is used everyday in all of my classes.

 

[killspray]

.

 

[OZ Dude]

Oh oh.

I can see a sin of trouble...

There's gonna be a showdown cos there's a difference of opinion

Someone is gonna end up with a tan on their a$$

Wow, that was really bad wasn't it but someone had to do it :fart:

Oz

 

[NVWSSV]

Math Sucks.

Edit: Because I suck at it.