Another Trig Identity


by cscott
Tags: identity, trig
cscott
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#1
Oct9-05, 01:55 PM
P: 786
[tex]1 - \frac{\cos^2 \theta}{1 + \sin^2 \theta} = \sin \theta[/tex]

I can only get to [tex] \frac{2\sin^2 \theta}{1 + \sin^2 \theta}[/tex] and I don't know if that's correct.
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TD
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#2
Oct9-05, 04:19 PM
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Perhaps you should check the problem again because it doesn't seem correct to me. Try [itex]\theta = 3\pi /2[/itex], the LHS will be 1 while the RHS is -1.
Fermat
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#3
Oct9-05, 04:22 PM
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It's not an identity!

Let θ = -π/2

then lhs = 1 - cosē(-π/2)/(1 + sinē(-π/2))
lhs = 1 - 0/2
lhs = 1
=====

but rhs = sin(-π/2) = -sin(π/2) = -1
rhs = - 1
=======

Since lhs ≠ rhs, it can't be an identity.

cscott
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#4
Oct9-05, 04:35 PM
P: 786

Another Trig Identity


OMG, I've spent so much time trying to figure it out... I never thought of checking to make sure it was indeed an identity. I guess my teacher made a mistake. Thanks guys.

(Don't worry, I'll be back with another too soon )
Fermat
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#5
Oct9-05, 04:45 PM
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I did a quick check to see if it was an identity. After all , you were having trouble !
It checked out ok for θ = 0 and θ = pi/2, so I did some work on it.
It was only later on that I plugged the expressions into graphmatica and found out they were actually two different functions !!
cscott
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#6
Oct10-05, 02:39 PM
P: 786
[tex]\frac{1 - \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 - \sin \theta} = 2 \sec \theta[/tex]

[tex]\frac{1 - \sin \theta}{1 + \sin \theta} = (\sec \theta - \tan \theta)^2[/tex]

I can't seem to get anywhere with these...
Fermat
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#7
Oct10-05, 04:25 PM
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P: 879
For the 1st one, cross-multiply the two terms in the lhs and simplify.
Fermat
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#8
Oct10-05, 04:30 PM
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P: 879
For the 2nd one, multiply the lhs by (1 - sinθ) and simplify.
cscott
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#9
Oct10-05, 07:26 PM
P: 786
I can only get to [itex]2 \sin \theta + 2 \sin^2 \theta[/itex] for the first after cross multiplying.

For the second, how can I just multiply by [itex](1 - \sin \theta)[/itex]? I thought you can only multiply by 1?
arildno
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#10
Oct10-05, 07:48 PM
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Quote Quote by cscott
[tex]\frac{1 - \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 - \sin \theta} = 2 \sec \theta[/tex]

[tex]\frac{1 - \sin \theta}{1 + \sin \theta} = (\sec \theta - \tan \theta)^2[/tex]

I can't seem to get anywhere with these...
[tex]\frac{1 - \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 - \sin \theta}=\frac{1 - \sin \theta}{\cos \theta}+\frac{1+\sin\theta}{1+\sin\theta}\frac{\cos\theta}{1-\sin\theta}=\frac{1 - \sin \theta}{\cos \theta} +\frac{1+\sin\theta}{\cos\theta}=\frac{2}{\cos\theta}=2sec\theta[/tex]

[tex]\frac{1 - \sin \theta}{1 + \sin \theta} =\frac{1-\sin\theta}{1-\sin\theta}\frac{1-\sin\theta}{1+\sin\theta}=\frac{(1-\sin\theta)^{2}}{\cos^{2}\theta}=(sec\theta-tan\theta)^{2}[/tex]
cscott
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#11
Oct10-05, 08:04 PM
P: 786
I see, thanks.
Fermat
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#12
Oct10-05, 11:34 PM
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Quote Quote by cscott
I can only get to [itex]2 \sin \theta + 2 \sin^2 \theta[/itex] for the first after cross multiplying.

For the second, how can I just multiply by [itex](1 - \sin \theta)[/itex]? I thought you can only multiply by 1?
Ah, sorry, that should have been, multiply above and below by (1 - sinθ)


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