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Trig identity

  • Thread starter clook
  • Start date
  • #1
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how can i prove

tan X/sinx+cosx=sin^2 X + sinXcosX/cos X - 2cos^3X

so far i've tried using basic identities to figure it out, but i just end up getting confused.


:confused:
 
Last edited:

Answers and Replies

  • #2
115
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The way you have written it is confusing. Are there any paranthesis missing?
 
  • #3
35
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assyrian_77 said:
The way you have written it is confusing. Are there any paranthesis missing?
no..

hmm lemme write it out another way

how can i prove

tanX
______
sinx+cosx

=

sin^2 X + sinXcosX
________________
cos X - 2cos^3X
 
  • #4
CarlB
Science Advisor
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Try putting x = pi/4 and see what you get.

Carl
 
  • #5
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In other words, there were parenthesis missing...
 
  • #6
VietDao29
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clook said:
no..

hmm lemme write it out another way

how can i prove

tanX
______
sinx+cosx

=

sin^2 X + sinXcosX
________________
cos X - 2cos^3X
The problem is wrong. Are you sure you copied this correctly?
I think it should be:
[tex]\frac{\tan x}{\sin x + \cos x} = \frac{\sin ^ 2 x + \sin x \cos x}{\cos x + 2 \sin x \cos ^ 2 x}[/tex]
 
  • #7
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VietDao29 said:
The problem is wrong. Are you sure you copied this correctly?
I think it should be:
[tex]\frac{\tan x}{\sin x + \cos x} = \frac{\sin ^ 2 x + \sin x \cos x}{\cos x + 2 \sin x \cos ^ 2 x}[/tex]
i copied it as it was exactly written in the book.. :confused:
 
  • #8
arildno
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clook said:
how can i prove

tan X/sinx+cosx=sin^2 X + sinXcosX/cos X - 2cos^3X

so far i've tried using basic identities to figure it out, but i just end up getting confused.


:confused:
No, you haven't copied this correctly: Put X=0
Left hand side yields 2, whereas right-hand side yields -2.

Read your book again; this time with your eyes open.
 
  • #9
335
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Try cancelling the common factors and try compreesing the question to its simplest form. You would then probably then get a breakthrough
 
  • #10
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arildno said:
No, you haven't copied this correctly: Put X=0
Left hand side yields 2, whereas right-hand side yields -2.

Read your book again; this time with your eyes open.
i am positive i copied it correctly. i would take a picture from the book, but i don't have a camera or scanner right now.
 
  • #11
Hootenanny
Staff Emeritus
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When you put x = 0 in both sides are equal to zero.

Just rewritting what clook posted in a more organised manor:

[tex]\frac{\tan x}{\sin x + \cos x} = \frac{\sin^2 x + \sin x \cos x}{\cos x - 2\cos^3 x}[/tex]
 
  • #12
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Hootenanny said:
When you put x = 0 in both sides are equal to zero.

Just rewritting what clook posted in a more organised manor:

[tex]\frac{\tan x}{\sin x + \cos x} = \frac{\sin^2 x + \sin x \cos x}{\cos x - 2\cos^3 x}[/tex]
This statement is equivalent to saying [tex]sinx=-cosx[/tex], so something is VERY fishy here.

-Dan
 
  • #13
115
0
Ok clook, assuming that

[tex]\frac{\tan x}{\sin x + \cos x} = \frac{\sin^2 x + \sin x \cos x}{\cos x - 2\cos^3 x}[/tex]

is what you have in your book. I worked with the RHS and got this:

[tex]\frac{\tan x}{\sin x-\cos x}[/tex]

As you can see, this is different from the LHS in your equation. Something is indeed fishy here.
 
  • #14
VietDao29
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clook said:
i am positive i copied it correctly. i would take a picture from the book, but i don't have a camera or scanner right now.
Okay, look at your problem again, does it say something like:
Prove this identity, or does it tell you to solve the equation?
 
  • #15
Hootenanny
Staff Emeritus
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VietDao29 said:
Okay, look at your problem again, does it say something like:
Prove this identity, or does it tell you to solve the equation?
Yes solve would make alot more sense (and would be very easy as well) :wink:
 

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