Fdtxjr

How do I free up internal storage if I don't have any apps downloaded?

Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?

How can I define good in a religion that claims no moral authority?

How do you keep chess fun when your opponent constantly beats you?

Getting crown tickets for Statue of Liberty

How to support a colleague who finds meetings extremely tiring?

Is an up-to-date browser secure on an out-of-date OS?

How to obtain a position of last non-zero element

Can I have a signal generator on while it's not connected?

What does Linus Torvalds mean when he says that Git "never ever" tracks a file?

How much of the clove should I use when using big garlic heads?

Why doesn't UInt have a toDouble()?

Can a flute soloist sit?

If a sorcerer casts the Banishment spell on a PC while in Avernus, does the PC return to their home plane?

Output the Arecibo Message

Is it okay to consider publishing in my first year of PhD?

Did Scotland spend $250,000 for the slogan "Welcome to Scotland"?

What's the name of these plastic connectors

Likelihood that a superbug or lethal virus could come from a landfill

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

How to notate time signature switching consistently every measure

A word that means fill it to the required quantity

Relationship between Gromov-Witten and Taubes' Gromov invariant

What do hard-Brexiteers want with respect to the Irish border?

Reaching upon $9=1$ while solving $x$ for $3tan(x-15^circ)=tan(x+15^circ)$

6

2

$begingroup$

$x$ for $3tan(x-15^circ)=tan(x+15^circ)$

Substituting $y=x+45^circ$, we get
$$3tan(y-60^circ)=tan(y-30^circ)$$
$$3fractan y - sqrt31+sqrt3tan y=fractan y - 1/sqrt31+1/sqrt3cdottan y$$

$$3(tan ^2 y-3)=3tan ^2-1$$

$$9=1$$

The solution provided by the book $x=npi + pi/4$ fits, so why did i get $9=1$?

trigonometry proof-explanation

share|cite|improve this question

edited Mar 24 at 5:43

Michael Rozenberg

110k1896201

asked Mar 24 at 5:17

user619072

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get to the third line ($3(tan^2 y-3) cdots$)?
  $endgroup$
  – Toby Mak
  Mar 24 at 5:28
  

  
  
  
  

add a comment | 

6

2

$begingroup$

$x$ for $3tan(x-15^circ)=tan(x+15^circ)$

Substituting $y=x+45^circ$, we get
$$3tan(y-60^circ)=tan(y-30^circ)$$
$$3fractan y - sqrt31+sqrt3tan y=fractan y - 1/sqrt31+1/sqrt3cdottan y$$

$$3(tan ^2 y-3)=3tan ^2-1$$

$$9=1$$

The solution provided by the book $x=npi + pi/4$ fits, so why did i get $9=1$?

trigonometry proof-explanation

share|cite|improve this question

edited Mar 24 at 5:43

Michael Rozenberg

110k1896201

asked Mar 24 at 5:17

user619072

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get to the third line ($3(tan^2 y-3) cdots$)?
  $endgroup$
  – Toby Mak
  Mar 24 at 5:28
  

  
  
  
  

add a comment | 

$begingroup$

$x$ for $3tan(x-15^circ)=tan(x+15^circ)$

Substituting $y=x+45^circ$, we get
$$3tan(y-60^circ)=tan(y-30^circ)$$
$$3fractan y - sqrt31+sqrt3tan y=fractan y - 1/sqrt31+1/sqrt3cdottan y$$

$$3(tan ^2 y-3)=3tan ^2-1$$

$$9=1$$

The solution provided by the book $x=npi + pi/4$ fits, so why did i get $9=1$?

trigonometry proof-explanation

share|cite|improve this question

edited Mar 24 at 5:43

Michael Rozenberg

110k1896201

asked Mar 24 at 5:17

user619072

$endgroup$

share|cite|improve this question

edited Mar 24 at 5:43

Michael Rozenberg

110k1896201

asked Mar 24 at 5:17

user619072

share|cite|improve this question

edited Mar 24 at 5:43

Michael Rozenberg

110k1896201

asked Mar 24 at 5:17

user619072

edited Mar 24 at 5:43

Michael Rozenberg

110k1896201

edited Mar 24 at 5:43

Michael Rozenberg

110k1896201

3 Answers
3

active

oldest

votes

4

$begingroup$

You got $$0cdottan^2y+9=1.$$ Id est, you got that $tany$ does not exist.

Thus, $$y=fracpi2+pi n,$$ where $ninmathbbZ,$ or
$$x+45^circ=fracpi2+pi n,$$ which gives $$x=pi n+fracpi4.$$
We did not get a contradiction in Math!

share|cite|improve this answer

edited Mar 24 at 5:47

answered Mar 24 at 5:33

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $9=1$, then how sure are u that it is $tan y$ that does not exist? It could be $tan (x-15)$ or the other trigonometric ratios...
  $endgroup$
  – rash
  Mar 24 at 5:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rash We got $0cdottan^2y+9=1.$
  $endgroup$
  – Michael Rozenberg
  Mar 24 at 5:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rash: Another way to look at it ... From the step $3 tan^2y-9 = 3 tan^2y - 1$, we see that subtracting either $9$ or $1$ from $3tan^2y$ gives the same result. We can (somewhat informally) conclude that $3tan^2y$ —and, more specifically, $tan y$— must be infinite.
  $endgroup$
  – Blue
  Mar 24 at 5:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One solution of X-9=X-1is X is undefined or X tends to infinity.
  $endgroup$
  – Tojrah
  Mar 24 at 6:01
  

  
  
  
  

add a comment | 

4

$begingroup$

Hint:

Set $tan y=dfrac1a$

$$dfrac3(1-sqrt3a)a+sqrt3=dfracsqrt3-asqrt3a+1$$

$$iff3-a^2=3(1-3a^2)iff a=0$$

share|cite|improve this answer

answered Mar 24 at 6:03

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

add a comment | 

3

$begingroup$

Another way to avoid confusion:

$$dfrac31=dfracsin(x+15^circ)cos(x-15^circ)cos(x+15^circ)sin(x-15^circ)$$

Apply Componendo et Dividendo

$$dfrac3+13-1=dfracsin2xsin30^circ$$

share|cite|improve this answer

answered Mar 24 at 5:57

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've never heard of Componendo et Dividendo! $(+1)$
  $endgroup$
  – user477343
  Mar 24 at 6:09
  
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160130%2freaching-upon-9-1-while-solving-x-for-3-tanx-15-circ-tanx15-ci%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

You got $$0cdottan^2y+9=1.$$ Id est, you got that $tany$ does not exist.

Thus, $$y=fracpi2+pi n,$$ where $ninmathbbZ,$ or
$$x+45^circ=fracpi2+pi n,$$ which gives $$x=pi n+fracpi4.$$
We did not get a contradiction in Math!

share|cite|improve this answer

edited Mar 24 at 5:47

answered Mar 24 at 5:33

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $9=1$, then how sure are u that it is $tan y$ that does not exist? It could be $tan (x-15)$ or the other trigonometric ratios...
  $endgroup$
  – rash
  Mar 24 at 5:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rash We got $0cdottan^2y+9=1.$
  $endgroup$
  – Michael Rozenberg
  Mar 24 at 5:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rash: Another way to look at it ... From the step $3 tan^2y-9 = 3 tan^2y - 1$, we see that subtracting either $9$ or $1$ from $3tan^2y$ gives the same result. We can (somewhat informally) conclude that $3tan^2y$ —and, more specifically, $tan y$— must be infinite.
  $endgroup$
  – Blue
  Mar 24 at 5:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One solution of X-9=X-1is X is undefined or X tends to infinity.
  $endgroup$
  – Tojrah
  Mar 24 at 6:01
  

  
  
  
  

add a comment | 

4

$begingroup$

You got $$0cdottan^2y+9=1.$$ Id est, you got that $tany$ does not exist.

Thus, $$y=fracpi2+pi n,$$ where $ninmathbbZ,$ or
$$x+45^circ=fracpi2+pi n,$$ which gives $$x=pi n+fracpi4.$$
We did not get a contradiction in Math!

share|cite|improve this answer

edited Mar 24 at 5:47

answered Mar 24 at 5:33

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $9=1$, then how sure are u that it is $tan y$ that does not exist? It could be $tan (x-15)$ or the other trigonometric ratios...
  $endgroup$
  – rash
  Mar 24 at 5:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rash We got $0cdottan^2y+9=1.$
  $endgroup$
  – Michael Rozenberg
  Mar 24 at 5:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @rash: Another way to look at it ... From the step $3 tan^2y-9 = 3 tan^2y - 1$, we see that subtracting either $9$ or $1$ from $3tan^2y$ gives the same result. We can (somewhat informally) conclude that $3tan^2y$ —and, more specifically, $tan y$— must be infinite.
  $endgroup$
  – Blue
  Mar 24 at 5:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One solution of X-9=X-1is X is undefined or X tends to infinity.
  $endgroup$
  – Tojrah
  Mar 24 at 6:01
  

  
  
  
  

add a comment | 

$begingroup$

You got $$0cdottan^2y+9=1.$$ Id est, you got that $tany$ does not exist.

Thus, $$y=fracpi2+pi n,$$ where $ninmathbbZ,$ or
$$x+45^circ=fracpi2+pi n,$$ which gives $$x=pi n+fracpi4.$$
We did not get a contradiction in Math!

share|cite|improve this answer

edited Mar 24 at 5:47

answered Mar 24 at 5:33

Michael RozenbergMichael Rozenberg

110k1896201

$endgroup$

share|cite|improve this answer

edited Mar 24 at 5:47

answered Mar 24 at 5:33

Michael RozenbergMichael Rozenberg

110k1896201

answered Mar 24 at 5:33

Michael RozenbergMichael Rozenberg

110k1896201

answered Mar 24 at 5:33

Michael RozenbergMichael Rozenberg

110k1896201

4

$begingroup$

Hint:

Set $tan y=dfrac1a$

$$dfrac3(1-sqrt3a)a+sqrt3=dfracsqrt3-asqrt3a+1$$

$$iff3-a^2=3(1-3a^2)iff a=0$$

share|cite|improve this answer

answered Mar 24 at 6:03

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

add a comment | 

4

$begingroup$

Hint:

Set $tan y=dfrac1a$

$$dfrac3(1-sqrt3a)a+sqrt3=dfracsqrt3-asqrt3a+1$$

$$iff3-a^2=3(1-3a^2)iff a=0$$

share|cite|improve this answer

answered Mar 24 at 6:03

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

add a comment | 

$begingroup$

Hint:

Set $tan y=dfrac1a$

$$dfrac3(1-sqrt3a)a+sqrt3=dfracsqrt3-asqrt3a+1$$

$$iff3-a^2=3(1-3a^2)iff a=0$$

share|cite|improve this answer

answered Mar 24 at 6:03

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

share|cite|improve this answer

answered Mar 24 at 6:03

lab bhattacharjeelab bhattacharjee

228k15159279

answered Mar 24 at 6:03

lab bhattacharjeelab bhattacharjee

228k15159279

answered Mar 24 at 6:03

lab bhattacharjeelab bhattacharjee

228k15159279

3

$begingroup$

Another way to avoid confusion:

$$dfrac31=dfracsin(x+15^circ)cos(x-15^circ)cos(x+15^circ)sin(x-15^circ)$$

Apply Componendo et Dividendo

$$dfrac3+13-1=dfracsin2xsin30^circ$$

share|cite|improve this answer

answered Mar 24 at 5:57

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've never heard of Componendo et Dividendo! $(+1)$
  $endgroup$
  – user477343
  Mar 24 at 6:09
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Another way to avoid confusion:

$$dfrac31=dfracsin(x+15^circ)cos(x-15^circ)cos(x+15^circ)sin(x-15^circ)$$

Apply Componendo et Dividendo

$$dfrac3+13-1=dfracsin2xsin30^circ$$

share|cite|improve this answer

answered Mar 24 at 5:57

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've never heard of Componendo et Dividendo! $(+1)$
  $endgroup$
  – user477343
  Mar 24 at 6:09
  
  

  
  
  
  

add a comment | 

$begingroup$

Another way to avoid confusion:

$$dfrac31=dfracsin(x+15^circ)cos(x-15^circ)cos(x+15^circ)sin(x-15^circ)$$

Apply Componendo et Dividendo

$$dfrac3+13-1=dfracsin2xsin30^circ$$

share|cite|improve this answer

answered Mar 24 at 5:57

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

share|cite|improve this answer

answered Mar 24 at 5:57

lab bhattacharjeelab bhattacharjee

228k15159279

answered Mar 24 at 5:57

lab bhattacharjeelab bhattacharjee

228k15159279

answered Mar 24 at 5:57

lab bhattacharjeelab bhattacharjee

228k15159279