Verify Identity: cosx - cosx/1 - tanx = sinx cosx/sinx -...

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KR3W

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Hello, I am having extensive trouble with Analytical Trigonometry. It's just really difficult for me. I'm fairly good at math, but this part of Trigonometry troubles me.
Anyways, if you guys can help me attack problems in a simple way, I'd appreciate it.

Here's one of the problems in my homework:

cosx - cosx/ 1 - tanx = sinx cosx/ sinx - cosx



Thank You. :)
 
Re: Verify The Identity

KR3W said:
Hello, I am having extensive trouble with Analytical Trigonometry. It's just really difficult for me. I'm fairly good at math, but this part of Trigonometry troubles me.
Anyways, if you guys can help me attack problems in a simple way, I'd appreciate it.

Here's one of the problems in my homework:

cosx - cosx/ 1 - tanx = sinx cosx/ sinx - cosx



Thank You. :)
Click to expand...

substitute tan(x) = sin(x)/cos(x), multiply numerator and denominator by cos(x) and then include the first cos(x) term to that fraction by making one fraction.
 
Firstly, reduce the whole left-hand expression into a single fraction:

\(\displaystyle cos(x) - \frac{cos(x)}{1-tan(x)} = \frac{cos(x)(1-tan(x))-cos(x)}{1-tan(x)} = \frac{-cos(x)tan(x)}{1-tan(x)} = \frac{tan(x)cos(x)}{tan(x)-1}\)

Then multiply the numerator and denominator by cos(x):

\(\displaystyle \frac{tan(x)cos(x)}{tan(x)-1} = \frac{sin(x)cos(x)}{sin(x)-cos(x)}\)

There you have your original identity. Don't forget to replace the equals signs with identity signs, and that:

\(\displaystyle tan(x) = \frac{sin(x)}{cos(x)}\)
 
Trebor said:
Firstly, reduce the whole left-hand expression into a single fraction:

\(\displaystyle cos(x) - \frac{cos(x)}{1-tan(x)} = \frac{cos(x)(1-tan(x))-cos(x)}{1-tan(x)} = \frac{-cos(x)tan(x)}{1-tan(x)} = \frac{tan(x)cos(x)}{tan(x)-1}\)

Then multiply the numerator and denominator by cos(x):

\(\displaystyle \frac{tan(x)cos(x)}{tan(x)-1} = \frac{sin(x)cos(x)}{sin(x)-cos(x)}\)

There you have your original identity. Don't forget to replace the equals signs with identity signs, and that:

\(\displaystyle tan(x) = \frac{sin(x)}{cos(x)}\)
Click to expand...
Thanks for your help, as well as the other poster who helped me; but I understood everything up until you said to "multiply the numerator and denominator by cos (x)"
Can you help me with that? :)
 
Re: Verify Identity: cosx - cosx/1 - tanx = sinx cosx/sinx -

Hello, KR3W!

Here's the way I'd approach it . . .

\(\displaystyle \cos x\,-\,\L\frac{cos x}{1\,-\,\tan x}\:=\:\frac{\sin x\cdot\cos x}{\sin x\,-\,\cos x}\)
Click to expand...

On the left we have: \(\displaystyle \:cos x \;-\;\L\frac{\cos x}{1\,-\,\frac{\sin x}{\cos x}}\)

Multiply top and bottom by \(\displaystyle cos x:\) \(\displaystyle \;\cos x \,-\,\frac{\cos x(\cos x)}{\cos x\left(1 \,-\,\L\frac{\sin x}{\cos x}\right)}\;\) **

. . \(\displaystyle =\;\cos x\,-\,\L\frac{\cos^2x}{\cos x\,-\,\sin x} \;=\;\frac{\cos x(\cos x\,-\,\sin x)\,-\,\cos^2x}{\cos x\,-\,\sin x}\)

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

Multiply top and bottom by \(\displaystyle \,-1\):

. . \(\displaystyle \L\frac{-1(-\sin x\cdot\cos x)}{-1(\cos x\,-\,\sin x)} \;=\;\frac{\sin x\cdot\cos x}{\sin x\,-\,\cos x}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

This is similar to Trebor's "magic step".

He had: \(\displaystyle \L\:\frac{\tan x\cdot\cos x}{\tan x\,-\,1} \;=\;\frac{\frac{\sin x}{\cos x}\cdot\cos x}{\frac{\sin x}{\cos x}\,-\,1}\)


Multiply top and bottom by \(\displaystyle cos x:\)

. . \(\displaystyle \L\frac{\cos x\left(\frac{\sin x}{\cos x}\cdot\cos x\right)}{\cos x\left(\frac{\sin x}{\cos x} \,-\,1\right)} \;=\;\frac{\sin x\cdot\cos x}{\sin x\,-\,\cos x}\)

See how it works?

 
Thanks alot. I really appreciate it. I'm beginning to understand it really well. :)
 
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