Trig Identity


by cscott
Tags: identity, trig
cscott
cscott is offline
#1
Oct8-05, 12:05 PM
P: 786
Can someone please help me establish this identity?

[tex]\cos \theta (\tan \theta + \cot \theta) = \csc \theta[/tex]
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irony of truth
irony of truth is offline
#2
Oct8-05, 12:19 PM
P: 93
So, are you proving this identity?

Express your tangent and cotangent in terms of sine and cosine. Get their LCD... and your numerator becomes a well-known trigonometric identity..

Can you continue from here? :D
cscott
cscott is offline
#3
Oct8-05, 12:23 PM
P: 786
Quote Quote by irony of truth
So, are you proving this identity?

Express your tangent and cotangent in terms of sine and cosine. Get their LCD... and your numerator becomes a well-known trigonometric identity..

Can you continue from here? :D
The easy ones always get me :\

Thanks!

cscott
cscott is offline
#4
Oct8-05, 04:54 PM
P: 786

Trig Identity


I can't get this one either:

[tex]\frac{1 + \tan \theta}{1 - \tan \theta} = \frac{\cot \theta + 1}{\cot \theta - 1}[/tex]

I'm so bad at proofs
mezarashi
mezarashi is offline
#5
Oct8-05, 04:58 PM
HW Helper
P: 661
For this one, you can either choose to replace tan x by 1/cot x or replace cot x by 1/tan x. Choose either and do some algebriac manipulations while leaving the other side alone.
TD
TD is offline
#6
Oct8-05, 04:59 PM
HW Helper
P: 1,024
Or, if that doesn't work for you, substitute tan by sin/cos and cot by cos/sin, then simplify the expressions

Try, if you get stuck, show us!
cscott
cscott is offline
#7
Oct8-05, 05:16 PM
P: 786
I end up with

[tex]\frac{\cos^2 \theta + \sin \theta \cos \theta}{\cos^2 \theta - \sin \theta \cos \theta}[/tex]

or

[tex]\frac{\cot^2 \theta + \cot \theta}{\cot^2 \theta - \cot \theta}[/tex]

How do I continue?
TD
TD is offline
#8
Oct8-05, 05:22 PM
HW Helper
P: 1,024
How did you end up with that?

For the LHS:

[tex]\frac{{1 + \tan \theta }}{{1 - \tan \theta }} = \frac{{1 + \frac{{\sin \theta }}{{\cos \theta }}}}{{1 - \frac{{\sin \theta }}{{\cos \theta }}}} = \frac{{\frac{{\cos \theta + \sin \theta }}{{\cos \theta }}}}{{\frac{{\cos \theta - \sin \theta }}{{\cos \theta }}}} = \frac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}[/tex]

Now try the RHS
cscott
cscott is offline
#9
Oct8-05, 05:30 PM
P: 786
Quote Quote by TD
How did you end up with that?

For the LHS:

[tex]\frac{{1 + \tan \theta }}{{1 - \tan \theta }} = \frac{{1 + \frac{{\sin \theta }}{{\cos \theta }}}}{{1 - \frac{{\sin \theta }}{{\cos \theta }}}} = \frac{{\frac{{\cos \theta + \sin \theta }}{{\cos \theta }}}}{{\frac{{\cos \theta - \sin \theta }}{{\cos \theta }}}} = \frac{{\cos \theta + \sin \theta }}{{\cos \theta - \sin \theta }}[/tex]

Now try the RHS
Silly me - I just multiplied out the numerator by the reciprocal of the denomenator instead of just canceling out the cosines. If you factor the top and bottom of my expression you end up with what your answer. If I do this using 1/cot = tan I end up with the RHS.

Don't I need to continue with the LHS until I get the right or vice versa?
TD
TD is offline
#10
Oct8-05, 05:31 PM
HW Helper
P: 1,024
Well now you have the LHS, the easiest would be trying to get the same starting with the RHS, which will go more or less the same
cscott
cscott is offline
#11
Oct8-05, 05:34 PM
P: 786
Ah, I see. Thank you both of you.
TD
TD is offline
#12
Oct8-05, 05:35 PM
HW Helper
P: 1,024
No problem


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