Verify This Trigonometric Identity

Feb 2012
9
0
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!
 

The Chaz

Forum Staff
Nov 2009
2,767
5
Northwest Arkansas
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).
 
Feb 2012
9
0
Thanks for your reply. For some reason my professor wants us to work ONLY with one side. I'll give it a shot!
 

soroban

Math Team
Dec 2006
3,267
408
Lexington, MA
Hello, reaper!

This one required some Olympic-level gymnastics . . .


\(\displaystyle \text{Verify: }\: \frac{ \cot x\cdot\cos x}{\cot x + \cos x} = \frac{\cot x - \cos x}{\cot x\cdot\cos x}\)

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

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

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


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

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

 
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The Chaz

Forum Staff
Nov 2009
2,767
5
Northwest Arkansas
...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.
 
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Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
Here's another approach:

\(\displaystyle \frac{\cot x\cos x}{\cot x+\cos x} = \frac{\cot x\cos x}{(\csc x+1)\cos x}\cdot\frac{\csc x-1}{\csc x-1} = \frac{\cot x(\cot x-\cos x)}{\cot^2x\cos x} = \frac{\cot x-\cos x}{\cot x\cos x}\)
 
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Feb 2012
9
0
Hello, reaper!

This one required some Olympic-level gymnastics . . .


\(\displaystyle \text{Verify: }\: \frac{ \cot x\cdot\cos x}{\cot x + \cos x} = \frac{\cot x - \cos x}{\cot x\cdot\cos x}\)

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

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

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


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

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

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

By the way, you guys rock!
 
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skipjack

Forum Staff
Dec 2006
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\(\displaystyle \text{LHS} = \frac{\cot x\cos x}{\cot x\cos x(\sec x + \tan x)}\cdot\frac{\sec x - \tan x}{\sec x - \tan x} = \frac{\cot x - \cos x}{\cot x \cos x(\sec^2\!x - \tan^2\!x)}= \text{RHS}\)
 

greg1313

Forum Staff
Oct 2008
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London, Ontario, Canada - The Forest City
\(\displaystyle \frac{\cot x\cos x}{\cot x+\cos x}\cdot\frac{\tan x}{\tan x}=\frac{\cos x}{1+\sin x}=\frac{1-\sin x}{\cos x}\cdot\frac{\cot x}{\cot x}=\frac{\cot x-\cos x}{\cot x\cos x}\)
 

skipjack

Forum Staff
Dec 2006
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2,410
\(\displaystyle \frac{\cot x\cos x}{\cot x + \cos x}\cdot\frac{\cot x - \cos x}{\cot x - \cos x} = \frac{\cot x\cos x(\cot x - \cos x)}{\cot^2x - \cos^2x} = \frac{\cot x\cos x(\cot x - \cos x)}{(\csc^2x - 1)\cos^2x} = \frac{\cot x - \cos x}{\cot x\cos x}\)