anonymous
  • anonymous
Verify the trigonometric equation cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x)?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
SolomonZelman
  • SolomonZelman
are you sure it is not tan^2x ?
anonymous
  • anonymous
1 second please let me make sure

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SolomonZelman
  • SolomonZelman
oh, no nvm, it should be like that... but still make sure.
anonymous
  • anonymous
it is how I initially wrote it
SolomonZelman
  • SolomonZelman
I would start from multiplying times tan(x) on both sides.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x\sec^4x=\cot x+2\tan x+\tan^3x }\) \(\large\color{black}{ \displaystyle \sec^4x=1+2\tan^2x+\tan^4x }\)
anonymous
  • anonymous
I have to focus on one side though, I am trying to prove that they are the same, not solve.
SolomonZelman
  • SolomonZelman
oh, okay, it would have been wonderful, but I guess not
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\) would you agree with this?
SolomonZelman
  • SolomonZelman
I factored out of cot(x).
anonymous
  • anonymous
I have seen that everywhere I have looked, but I dont understand where the tan^2 or the Tan^4 come from
SolomonZelman
  • SolomonZelman
well, do you agree that: \(\cot(x) \times \tan(x)=1\) \(\cot(x) \times \tan^2(x)=\tan(x)\) \(\cot(x) \times \tan^3(x)=\tan^2(x)\) \(\cot(x) \times \tan^4(x)=\tan^3(x)\) and so on... \(\cot(x) \times \tan^{N}(x)=\tan^{N-1}(x)\)
SolomonZelman
  • SolomonZelman
if you don't understand why these are true, then ask
anonymous
  • anonymous
I didn't know those identities. I knew that sin^2x + cos^2x =1 ....
SolomonZelman
  • SolomonZelman
well, do you understand why they are logically true?
dinamix
  • dinamix
@SolomonZelman i want ask u how did u get sec^4(x) = 1 +2tan^2(x) +tan^4(x)
anonymous
  • anonymous
Not really.
SolomonZelman
  • SolomonZelman
I didn't get that yet, but we will get that.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle (x+1)^2=x^2+2x+1 }\) right?
dinamix
  • dinamix
we have only sec(x) = 1/cos(x)
SolomonZelman
  • SolomonZelman
turmoilx2, do you agree that: \(\cot(x) \times \tan(x)=1\)
anonymous
  • anonymous
I suppose... if you tell me that it's true
anonymous
  • anonymous
@dinamix . Are you in FLVS?
SolomonZelman
  • SolomonZelman
\(\cot(x) \times \tan(x)=1\) \(\dfrac{\cos(x)}{\sin(x)}\times \dfrac{\sin(x)}{\cos(x)}=1\)
anonymous
  • anonymous
I understand that.
SolomonZelman
  • SolomonZelman
turmolix, how about now?
anonymous
  • anonymous
Yes!
SolomonZelman
  • SolomonZelman
ok, my comp glitched:)
SolomonZelman
  • SolomonZelman
\(\cot(x) \times \tan(x)=1\) So, now, we will be multiplying times tan(x) on both sides: \(\cot(x) \times \tan^2(x)=\tan(x)\) \(\cot(x) \times \tan^3(x)=\tan^2(x)\) \(\cot(x) \times \tan^4(x)=\tan^3(x)\) and so on... \(\cot(x) \times \tan^{N}(x)=\tan^{N-1}(x)\) -------------------------------- Is this better now, why the list of these identites is true?
dinamix
  • dinamix
@turmoilx2 no mate
SolomonZelman
  • SolomonZelman
turmoilx, when you read, reply me if you got what I elaborates so far in my last reply.
anonymous
  • anonymous
I understand why the identities are true now
SolomonZelman
  • SolomonZelman
Ok, so now back to our question;)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\) Now, using these identies, I am factoring the left side out of \(\cot(x)\). \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\) Do you get how I am factoring?
anonymous
  • anonymous
No, please explain.
SolomonZelman
  • SolomonZelman
let me do an intermediate step in blue: \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\) \(\large\color{blue}{ \displaystyle \cot x \sec^4x=\cot x\cdot 1+2\cot x\tan^2x+\cot x\tan^4x }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\) is it better now?
SolomonZelman
  • SolomonZelman
after the blue, I factor out of \(\cot(x)\). If you still have questions about how I factored, then please ask.
anonymous
  • anonymous
Im still lost. Sorry if I'm being slow here, online math classes are not for me.
SolomonZelman
  • SolomonZelman
(it is fine if you don't I can attempt to explain a little better than what I did. I am not a very good tutor after all).
anonymous
  • anonymous
I dont understand how you can just multiply the tan by cot and I dont know how to get the exponents for the tangents still. I understand the identities you mentioned above but I dont know how they apply to this first step if that makes sense...
anonymous
  • anonymous
oh wait...
anonymous
  • anonymous
Is it allowed because Cotxtanx is the same as multiplying by 1 which wouldnt change the equation?
SolomonZelman
  • SolomonZelman
yes
anonymous
  • anonymous
I think I understand some more now...
SolomonZelman
  • SolomonZelman
but I would still just in case post my explanation on that as throuhg as I can. I will try not to take too much time.
anonymous
  • anonymous
Yes. That would be appreciated, thank you!
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\tan x}+\color{red}{\tan^3x} }\) `-----------------------------------------------`\({\\[0.8em]}\) \(\large\color{blue}{ \displaystyle \cot x =\cot x }\) (We know that)\({\\[0.8em]}\) `-----------------------------------------------`\({\\[0.8em]}\) \(\large\color{green}{ \displaystyle \tan x =\cot x\tan^2x }\)\({\\[0.8em]}\) (As we showed before in the list of those identites)\({\\[0.8em]}\) \(\large\color{green}{ \displaystyle \tan x =\cot x\tan^2x }\)\({\\[0.8em]}\) And then multiply times 2 on both sides\({\\[0.8em]}\) \(\large\color{green}{ \displaystyle 2\tan x =2\cot x\tan^2x }\) \({\\[0.8em]}\) `-----------------------------------------------`\({\\[0.8em]}\) \(\large\color{red}{ \displaystyle \tan^3x =\cot x\tan^4x }\)\({\\[0.8em]}\) (Also, as we showed before in the list of those identites)\({\\[0.8em]}\) `-----------------------------------------------` So it comes out that: \(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\cot x\tan^2 x}+\color{red}{\cot x\tan^4x} }\)
SolomonZelman
  • SolomonZelman
take your time.
SolomonZelman
  • SolomonZelman
When you are done reading, tell me if you understand how I got the following equation below: \(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\cot x\tan^2 x}+\color{red}{\cot x\tan^4x} }\)
anonymous
  • anonymous
Here is what I have so far: \[CotxSec^4x=Cotx+2Tanx+Tan^3x\] \[CotxSec^4x=Cotx+2(CotxTan^2x)+(CotxTan^4x)\] \[CotxSec^4x=Cotx+Cot(2Tan^2x+tan^4x) Correct???
anonymous
  • anonymous
oops \[CotxSec^4x=Cotx+Cot(2Tan^2x+tan^4x)\]
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}\) Rather, take out \(\cot(x)\) from the entire right side. \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\)
SolomonZelman
  • SolomonZelman
did you get everything I have done so far/
anonymous
  • anonymous
\[cotxsec4x=cotx((((1))))+2\tan2x+\tan4x)\] So Im assuming this "1" is from removing the cotagents... but how?
SolomonZelman
  • SolomonZelman
well when I factor cot(x) out of cot(x), then I get 1.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}\) I am factoring out of \(\cot(x)\). And I factor every term on the left side. \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\) still questions about that factoring out?
anonymous
  • anonymous
I think what I dont get is why Cotx+2CotxTan^2x doesnt become 1+2Tanx instead of 1+ 2Tan^2x
SolomonZelman
  • SolomonZelman
2tan^2x is what I got, didn't I?
SolomonZelman
  • SolomonZelman
oh, because you have 2cot(x)tan²(x), and thus when you take out the cot(x) part, you remain with 2tan²(x).
anonymous
  • anonymous
I'll just go with it. Cot can be factored from itself to equal 1.
SolomonZelman
  • SolomonZelman
yes, true
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\) then, you know that: \(\color{red}{({\rm w}+1)^2={\rm w}^2+2{\rm w}+1}\) So, then look at the parenthesis on the right side and tell me how THAT can be factors.
SolomonZelman
  • SolomonZelman
\(\large (\)\(\color{red}{\rm w}\) in our case is \(\tan^2x\)\(\large )\)
anonymous
  • anonymous
\[(w+1)^2=w^2+1^2\] I know that this much is true.
SolomonZelman
  • SolomonZelman
no, that is not true
SolomonZelman
  • SolomonZelman
w²+1 would be w²+1², BUT (w+1)² is different:
SolomonZelman
  • SolomonZelman
\((w+1)^2=(w+1)(w+1)=w(w+1)+1(w+1) \\ =w^2+w+w+1=w^2+2w+1\)
SolomonZelman
  • SolomonZelman
so, therefore we know that: \((w+1)^2=w^2+2w+1\)
SolomonZelman
  • SolomonZelman
ok, tell me if you get my very last post...
anonymous
  • anonymous
I get it now. Mind blown
anonymous
  • anonymous
You are very patient. Thanks
SolomonZelman
  • SolomonZelman
Ok, so now if you look at, and understand the fact that: \((w+1)^2=w^2+2w+1\) Then you should also know that: \((\tan^2x+1)^2=\left(\tan^2x\right)^2+2\left(\tan^2x\right)+1\) `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` And then take a look at your problem: \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\) and it should follow that: \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+\tan^2 x\right)^2}\)
SolomonZelman
  • SolomonZelman
By the way, it is not a problem.
anonymous
  • anonymous
and then (1+tan^2x)^2 = (sec^2x)^2 which then equals sec^4x
SolomonZelman
  • SolomonZelman
Yes, that is correct!
anonymous
  • anonymous
THANK YOU SO MUCH!
SolomonZelman
  • SolomonZelman
And then, we have verified the original equation to be an identity.
SolomonZelman
  • SolomonZelman
You are welcome:)

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