Trjtdtk

Why was the term "discrete" used in discrete logarithm?

Why didn't Eitri join the fight?

Why do people hide their license plates in the EU?

How does debian/ubuntu knows a package has a updated version

Remainder theorem for polynomials (JUEE 1990)

How do pianists reach extremely loud dynamics?

Why light coming from distant stars is not discrete?

Do I really need recursive chmod to restrict access to a folder?

Is it a good idea to use CNN to classify 1D signal?

Is it true that "carbohydrates are of no use for the basal metabolic need"?

Adverb for when you're not exaggerating

Can an alien society believe that their star system is the universe?

What does the "x" in "x86" represent?

How to deal with a team lead who never gives me credit?

How to override model in magento2?

When were vectors invented?

tcolorbox: Potential bug with duplicate label for hyperref link

How does the particle を relate to the verb 行く in the structure「A を + B に行く」?

Delete nth line from bottom

3 doors, three guards, one stone

A binary hook-length formula?

Why aren't air breathing engines used as small first stages

The logistics of corpse disposal

How to find out what spells would be useless to a blind NPC spellcaster?

Simplify trigonometric expression using trigonometric identities

3

$begingroup$

I have the trigonometric expression: $$2sin x +2sin left(fracpi 3 -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?

trigonometry

share|cite|improve this question

edited Apr 1 at 15:55

MarianD

2,2761618

asked Apr 1 at 15:19

zaz9999zaz9999

162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  mathworld.wolfram.com/ProsthaphaeresisFormulas.html
  $endgroup$
  – lab bhattacharjee
  Apr 1 at 15:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just between us: IMO the two expressions are of the same complexity.
  $endgroup$
  – Yves Daoust
  Apr 1 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  = 2 cos(x-π/6).
  $endgroup$
  – Anton Sherwood
  Apr 1 at 22:26
  

  
  
  
  

add a comment | 

3

$begingroup$

I have the trigonometric expression: $$2sin x +2sin left(fracpi 3 -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?

trigonometry

share|cite|improve this question

edited Apr 1 at 15:55

MarianD

2,2761618

asked Apr 1 at 15:19

zaz9999zaz9999

162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  mathworld.wolfram.com/ProsthaphaeresisFormulas.html
  $endgroup$
  – lab bhattacharjee
  Apr 1 at 15:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just between us: IMO the two expressions are of the same complexity.
  $endgroup$
  – Yves Daoust
  Apr 1 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  = 2 cos(x-π/6).
  $endgroup$
  – Anton Sherwood
  Apr 1 at 22:26
  

  
  
  
  

add a comment | 

$begingroup$

I have the trigonometric expression: $$2sin x +2sin left(fracpi 3 -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?

trigonometry

share|cite|improve this question

edited Apr 1 at 15:55

MarianD

2,2761618

asked Apr 1 at 15:19

zaz9999zaz9999

162

$endgroup$

share|cite|improve this question

edited Apr 1 at 15:55

MarianD

2,2761618

asked Apr 1 at 15:19

zaz9999zaz9999

162

share|cite|improve this question

edited Apr 1 at 15:55

MarianD

2,2761618

asked Apr 1 at 15:19

zaz9999zaz9999

162

edited Apr 1 at 15:55

MarianD

2,2761618

edited Apr 1 at 15:55

MarianD

2,2761618

asked Apr 1 at 15:19

zaz9999zaz9999

162

asked Apr 1 at 15:19

zaz9999zaz9999

162

3 Answers
3

active

oldest

votes

7

$begingroup$

Using the formula for $sin (alpha - beta)$ you obtain

beginalign
2&sin x +2sin left(fracpi 3 -xright)\
= 2&sin x +2left[sin left(fracpi 3right) cos x - cosleft(fracpi 3right) sin xright]\
= 2&sin x + 2left[sqrt 3 over 2 cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &colorredsin x + sqrt 3 cos x
endalign

share|cite|improve this answer

edited Apr 1 at 15:59

answered Apr 1 at 15:44

MarianDMarianD

2,2761618

$endgroup$

add a comment | 

4

$begingroup$

Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$

share|cite|improve this answer

answered Apr 1 at 15:26

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.1k42867

$endgroup$

add a comment | 

1

$begingroup$

Hint: $sin(fracpi3-x)=sin fracpi3cos x-cos fracpi3sin x$

share|cite|improve this answer

answered Apr 1 at 15:25

VasyaVasya

4,4771618

$endgroup$

add a comment | 

Your Answer

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%2f3170733%2fsimplify-trigonometric-expression-using-trigonometric-identities%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

7

$begingroup$

Using the formula for $sin (alpha - beta)$ you obtain

beginalign
2&sin x +2sin left(fracpi 3 -xright)\
= 2&sin x +2left[sin left(fracpi 3right) cos x - cosleft(fracpi 3right) sin xright]\
= 2&sin x + 2left[sqrt 3 over 2 cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &colorredsin x + sqrt 3 cos x
endalign

share|cite|improve this answer

edited Apr 1 at 15:59

answered Apr 1 at 15:44

MarianDMarianD

2,2761618

$endgroup$

add a comment | 

7

$begingroup$

Using the formula for $sin (alpha - beta)$ you obtain

beginalign
2&sin x +2sin left(fracpi 3 -xright)\
= 2&sin x +2left[sin left(fracpi 3right) cos x - cosleft(fracpi 3right) sin xright]\
= 2&sin x + 2left[sqrt 3 over 2 cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &colorredsin x + sqrt 3 cos x
endalign

share|cite|improve this answer

edited Apr 1 at 15:59

answered Apr 1 at 15:44

MarianDMarianD

2,2761618

$endgroup$

add a comment | 

$begingroup$

Using the formula for $sin (alpha - beta)$ you obtain

beginalign
2&sin x +2sin left(fracpi 3 -xright)\
= 2&sin x +2left[sin left(fracpi 3right) cos x - cosleft(fracpi 3right) sin xright]\
= 2&sin x + 2left[sqrt 3 over 2 cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &colorredsin x + sqrt 3 cos x
endalign

share|cite|improve this answer

edited Apr 1 at 15:59

answered Apr 1 at 15:44

MarianDMarianD

2,2761618

$endgroup$

share|cite|improve this answer

edited Apr 1 at 15:59

answered Apr 1 at 15:44

MarianDMarianD

2,2761618

answered Apr 1 at 15:44

MarianDMarianD

2,2761618

answered Apr 1 at 15:44

MarianDMarianD

2,2761618

4

$begingroup$

Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$

share|cite|improve this answer

answered Apr 1 at 15:26

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.1k42867

$endgroup$

add a comment | 

4

$begingroup$

Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$

share|cite|improve this answer

answered Apr 1 at 15:26

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.1k42867

$endgroup$

add a comment | 

$begingroup$

Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$

share|cite|improve this answer

answered Apr 1 at 15:26

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.1k42867

$endgroup$

share|cite|improve this answer

answered Apr 1 at 15:26

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.1k42867

answered Apr 1 at 15:26

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.1k42867

answered Apr 1 at 15:26

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

79.1k42867

1

$begingroup$

Hint: $sin(fracpi3-x)=sin fracpi3cos x-cos fracpi3sin x$

share|cite|improve this answer

answered Apr 1 at 15:25

VasyaVasya

4,4771618

$endgroup$

add a comment | 

1

$begingroup$

Hint: $sin(fracpi3-x)=sin fracpi3cos x-cos fracpi3sin x$

share|cite|improve this answer

answered Apr 1 at 15:25

VasyaVasya

4,4771618

$endgroup$

add a comment | 

$begingroup$

Hint: $sin(fracpi3-x)=sin fracpi3cos x-cos fracpi3sin x$

share|cite|improve this answer

answered Apr 1 at 15:25

VasyaVasya

4,4771618

$endgroup$

share|cite|improve this answer

answered Apr 1 at 15:25

VasyaVasya

4,4771618

answered Apr 1 at 15:25

VasyaVasya

4,4771618

answered Apr 1 at 15:25

VasyaVasya

4,4771618