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February 12th, 2012, 04:25 PM   #1
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Verify This Trigonometric Identity

For some reason, I can't seem to figure out how to verify the following using the fundamental identities.

( cot(x) * cos(x) )/( cot(x) + cos(x) ) = ( cot(x) - cos(x) )/( cot(x) * cos(x) )

I've tried multiplying by conjugates but that hasn't helped me, unless I'm doing something wrong.

Even if some one pushes me in the right direction, would be a help

Sorry for not making it look all nice with latex, but I don't know it well enough yet!
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February 12th, 2012, 05:03 PM   #2
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Re: Verify This Trigonometric Identity

Here's the most important tip, for all such problems: WORK ON BOTH SIDES!!!!!!!!!!!
Sure, you are usually required to work on just one side, but working on both sides is like putting one finger on the "start", and one finger on the "end" of a maze, and finding a way to get your fingers to touch. Once you've found that path in the maze, you can retrace it from start to finish. The same goes with trig identities.

So I would multiply both sides by tan/tan, using tan = sin/cos where necessary. This gives...
cos/(1 + sin) = (1 - sin)/cos

Cross multiplying gives the usual identity.

So can you find a way to work this on one side only?

The easiest way that I can think of would be to multiply the left by tan/tan, then use conjugates, then multiply by cot/cot (all on the left).
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February 12th, 2012, 05:21 PM   #3
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Re: Verify This Trigonometric Identity

Thanks for your reply. For some reason my professor wants us to work ONLY with one side. I'll give it a shot!
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February 12th, 2012, 05:24 PM   #4
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Re: Verify This Trigonometric Identity

Hello, reaper!

This one required some Olympic-level gymnastics . . .


Quote:
Originally Posted by reaper View Post
$\displaystyle \text{Verify: }\: \frac{ \cot x\cdot\cos x}{\cot x + \cos x} = \frac{\cot x - \cos x}{\cot x\cdot\cos x}$

$\displaystyle \text{We have: }\:\frac{\cot x\cdot\cos x}{\cot x + \cos x} = \frac{\frac{\cos x}{\sin x}\cdot\cos x}{\frac{\cos x}{\sin x} + \cos x}$

$\displaystyle \text{Multiply by }\frac{\sin x}{\sin x}:\;\;\frac{\sin x\left(\frac{\cos^{^2}x}{\sin x}\right)}{\sin x\left(\frac{\cos x}{\sin x} + \cos x\right)} = \frac{\cos^{^2}x}{\cos x + \sin x\cos x} = \frac{\cos^{^2}x}{\cos x(1 + \sin x)} = \frac{\cos x}{1 + \sin x}$

$\displaystyle \text{Multiply by }\frac{1-\sin x}{1-\sin x}:\;\;\frac{\cos x}{1 + \sin x}\,\cdot\,\frac{1 - \sin x}{1 - \sin x} = \frac{\cos x(1 - \sin x)}{1 - \sin^{^2}x} = \frac{\cos x(1 - \sin x)}{\cos^{^2}x}$


$\displaystyle \text{Divide numerator and denominator by }\sin x:$

$\displaystyle \frac{\frac{\cos x(1\,-\,\sin x)}{\sin x}}{\frac{\cos^2x}{\sin x}} \;=\;\frac{\frac{\cos x - \sin x\cos x}{\sin x}}{\frac{\cos^{^2}x}{\sin x}} = \frac{\frac{\cos x}{\sin x} - \cos x}{\frac{\cos x}{\sin x}\cdot\cos x} = \frac{\cot x - \cos x}{\cot x\cdot\cos x}$


Last edited by skipjack; August 26th, 2019 at 01:50 PM.
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February 12th, 2012, 05:29 PM   #5
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Re: Verify This Trigonometric Identity

Quote:
Originally Posted by reaper View Post
...For some reason my professor wants us to work ONLY with one side. I'll give it a shot!
That's par for the course, literally
I don't know why, exactly.

Last edited by skipjack; August 26th, 2019 at 01:51 PM.
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February 12th, 2012, 05:51 PM   #6
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Re: Verify This Trigonometric Identity

Here's another approach:

$\displaystyle \frac{\cot x\cos x}{\cot x+\cos x} = \frac{\cot x\cos x}{(\csc x+1)\cos x}\cdot\frac{\csc x-1}{\csc x-1} = \frac{\cot x(\cot x-\cos x)}{\cot^2x\cos x} = \frac{\cot x-\cos x}{\cot x\cos x}$

Last edited by skipjack; August 26th, 2019 at 01:33 PM.
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February 12th, 2012, 06:51 PM   #7
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Re: Verify This Trigonometric Identity

Quote:
Originally Posted by soroban View Post
Hello, reaper!

This one required some Olympic-level gymnastics . . .


Quote:
$\displaystyle \text{Verify: }\: \frac{ \cot x\cdot\cos x}{\cot x + \cos x} = \frac{\cot x - \cos x}{\cot x\cdot\cos x}$

$\displaystyle \text{We have: }\:\frac{\cot x\cdot\cos x}{\cot x + \cos x} = \frac{\frac{\cos x}{\sin x}\cdot\cos x}{\frac{\cos x}{\sin x} + \cos x}$

$\displaystyle \text{Multiply by }\frac{\sin x}{\sin x}:\;\;\frac{\sin x\left(\frac{\cos^{^2}x}{\sin x}\right)}{\sin x\left(\frac{\cos x}{\sin x} + \cos x\right)} = \frac{\cos^{^2}x}{\cos x + \sin x\cos x} = \frac{\cos^{^2}x}{\cos x(1 + \sin x)} = \frac{\cos x}{1 + \sin x}$

$\displaystyle \text{Multiply by }\frac{1-\sin x}{1-\sin x}:\;\;\frac{\cos x}{1 + \sin x}\,\cdot\,\frac{1 - \sin x}{1 - \sin x} = \frac{\cos x(1 - \sin x)}{1 - \sin^{^2}x} = \frac{\cos x(1 - \sin x)}{\cos^{^2}x}$


$\displaystyle \text{Divide numerator and denominator by }\sin x:$

$\displaystyle \frac{\frac{\cos x(1 - \sin x)}{\sin x}}{\frac{\cos^2x}{\sin x}} = \frac{\frac{\cos x - \sin x\cos x}{\sin x}}{\frac{\cos^{^2}x}{\sin x}} = \frac{\frac{\cos x}{\sin x} - \cos x}{\frac{\cos x}{\sin x}\cdot\cos x} = \frac{\cot x - \cos x}{\cot x\cdot\cos x}$

Just out of curiosity, why do you divide by sin x?

By the way, you guys rock!

Last edited by skipjack; August 26th, 2019 at 02:36 PM.
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August 26th, 2019, 02:33 PM   #8
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$\displaystyle \text{LHS} = \frac{\cot x\cos x}{\cot x\cos x(\sec x + \tan x)}\cdot\frac{\sec x - \tan x}{\sec x - \tan x} = \frac{\cot x - \cos x}{\cot x \cos x(\sec^2\!x - \tan^2\!x)}= \text{RHS}$
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August 26th, 2019, 08:41 PM   #9
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$\displaystyle \frac{\cot x\cos x}{\cot x+\cos x}\cdot\frac{\tan x}{\tan x}=\frac{\cos x}{1+\sin x}=\frac{1-\sin x}{\cos x}\cdot\frac{\cot x}{\cot x}=\frac{\cot x-\cos x}{\cot x\cos x}$
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August 26th, 2019, 10:37 PM   #10
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$\displaystyle \frac{\cot x\cos x}{\cot x + \cos x}\cdot\frac{\cot x - \cos x}{\cot x - \cos x} = \frac{\cot x\cos x(\cot x - \cos x)}{\cot^2x - \cos^2x} = \frac{\cot x\cos x(\cot x - \cos x)}{(\csc^2x - 1)\cos^2x} = \frac{\cot x - \cos x}{\cot x\cos x}$
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