Please help with proving Trig equality....

[AlphaWolf829]

0=theta
1-cos0 = sin0-1
1+cos0 sin0+1

need verification on what I think it is

 

[minendo]

wouldnt you just cross-multiply

 

[AlphaWolf829]

you can't when proving equalities.

 

[duragezic]

LOL I hate those damn things! I had a ton of trouble trying to do those and I posted a while back cause I needed a lot of help with them. I ended up understanding them fairly well, but that was a couple chapters and I have forgotten and I don't have my notes home.

Doesn't uh 1-cos = sin or is that only when its 1-cos^2 = sin^2?

 

[AlphaWolf829]

1-cos^2=sin^2 doesn't cos=1 over sin or something

 

[eLiu]

uh...that's not a real equality....

if theta=0

then u have (1-1)/(1+1)=(0-1)/(1), so 0=-1

not exactly true....

edit: oops, missed the fraction bar...my bad; same result though.

 

[eLiu]

<< 1-cos^2=sin^2 doesn't cos=1 over sin or something >>




that's quasi right....the first part is right--it's one of the pythagorean identities.

But, 1/sin is cosecant.

 

[AlphaWolf829]

no 0 is representing the greek letter theta since there is no specific angle

 

[eLiu]

<< no 0 is representing the greek letter theta since there is no specific angle >>



hmm? I'm saying if you set theta equal to 0. This is a condition for which your equality is not equal.

 

[AlphaWolf829]

ok thanks I though it didn't work I went through hours trying different ways to prove it but no luck I thought I was doing somethin wrong.

 

[Soccer55]

Multiply the left side by 1 using the conjugate: (1 - cos)/(1 - cos)
Then from there, it's just simplification and using the identity cos^2 + sin^2 = 1
Hope this helps.

-Tom

 

[safoo]

well eLiu is right that you have to restrict theta 0 < theta < pi/2 or the first quadrant.

also, i dont know y you said that you cant cross multiply when proving equalities.

heres wat i did to prove it wrong:

(1-cos0)(sin0+1) = (sin0-1)(1+cos0)

sin0+1 - cos0sin0 = sin0+cos0sin0 - 1 -cos0
2 - 2cos0sin0 = 0
2 - sin20 = 0 (by double angle formula sin20 = 2cos0sin0)

we know that sin20 can only be between 0 and 1, so 2-sin20 can never be 0 (or sin20 can never be 2)....

 

[RaynorWolfcastle]

what they said, that equality is incorrect as revealed by a simple cross multiplication

that's apart from the fact that at 180 and 270 degrees one of the sides of the equation is undefined

-Ice

 

[Soccer55]

<< what they said, that equality is incorrect as revealed by a simple cross multiplication >>


You can't solve these trig identities with cross multiplication. You have to manipulate the left side of the equality to get an expression equal to the right side or vice versa.



<< ...you have to restrict theta 0 < theta < pi/2 or the first quadrant >>


While I agree that there needs to be a restriction on theta, it seems like AlphaWolf829 just needs the method for simplifying the equality so the restriction doesn't really come into play.

-Tom

 

[safoo]

hmm...i cant seem to get it to be equal...hey alpha post the solution when u get it....

 

[RaynorWolfcastle]

<< what they said, that equality is incorrect as revealed by a simple cross multiplication >>


You can't solve these trig identities with cross multiplication. You have to manipulate the left side of the equality to get an expression equal to the right side or vice versa.

But you can check that the premise that you are working on is...



<< ...you have to restrict theta 0 < theta < pi/2 or the first quadrant >>


While I agree that there needs to be a restriction on theta, it seems like AlphaWolf829 just needs the method for simplifying the equality so the restriction doesn't really come into play.

the equality does not work at all, a quick and dirty maple graph proved that

-Ice

*edit* fixed quoting, I can post the graph if someone requests it

 

[Soccer55]

<< the equality does not work at all, a quick and dirty maple graph proved that >>



While it may not work with real values of theta, it does work algebraically. I didn't really look at it too closely, so you could very well be right that the equality fails in terms of actual usage. But it is definitely true using algebra and the trig identities.

-Tom

EDIT: I made a mistake in my work.....I'll post the solution once I fix it (assuming it can be fixed)

 

[RaynorWolfcastle]

errrr.... if it works in theory then it should work with "real values of theta", isn't the whole point of proving a trig equality being able to apply it?

-Ice

 

[Soccer55]

<< errrr.... if it works in theory then it should work with "real values of theta", isn't the whole point of proving a trig equality being able to apply it? >>



Yea, you're right.....I was thinking about say if it didn't work for theta = 90, but it did for every other value. It's been a long day After seeing my mistake in the work I did icecool83 is right, I don't think that there is any way to make it work (unless it was written wrong in the first place).

-Tom

 

[safoo]

ok thats good...for a minute there i thought i was losing my algebra skillz....

 

[Soccer55]

<< ok thats good...for a minute there i thought i was losing my algebra skillz >>



Nah.....that was me losing mine

-Tom

 

[RaynorWolfcastle]

<< Nah.....that was me losing mine >>


hey were all entitled to mistakes, it's good to hear that there are still some AT members that will admit to making them every so often :Q

a 10 rating goes to everyone in this thread for renewing my esteem in ATOT... one that had been severly lowered by a few users.

-Ice