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Another Trig Identity 
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#1
Oct905, 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. 


#2
Oct905, 04:19 PM

HW Helper
P: 1,021

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.



#3
Oct905, 04:22 PM

HW Helper
P: 876

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. 


#4
Oct905, 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 ) 


#5
Oct905, 04:45 PM

HW Helper
P: 876

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 !! 


#6
Oct1005, 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... 


#7
Oct1005, 04:25 PM

HW Helper
P: 876

For the 1st one, crossmultiply the two terms in the lhs and simplify.



#8
Oct1005, 04:30 PM

HW Helper
P: 876

For the 2nd one, multiply the lhs by (1  sinθ) and simplify.



#9
Oct1005, 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? 


#10
Oct1005, 07:48 PM

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P: 12,016

[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\thetatan\theta)^{2}[/tex] 


#11
Oct1005, 08:04 PM

P: 786

I see, thanks.



#12
Oct1005, 11:34 PM

HW Helper
P: 876




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