Xjyuk

Explain why a line can never intersect a plane in exactly two points.

Non-misogynistic way to say “asshole”?

Helping ease my back pain by studying 13 hours everyday , even weekends

Shooting someone's past self using special relativity

Does the monk's Step of the Wind feature activate the benefit of the Mobile feat?

Similarity score: Can Sklearn SVR predict values greater than 1 and less than 0?

Is it possible to get a mortgage with a custom duration in the US?

Why does the Saturn V have standalone inter-stage rings?

Dates on degrees don’t make sense – will people care?

Do I need a shock-proof watch for cycling?

Hit the Bulls Eye with T in the Center

Are all Ringwraiths called Nazgûl in LotR?

How to execute a command when ALL of the players are close enough

Pythagorean trigonometry identity

career in signal processing

Do I have to explain the mechanical superiority of the player-character within the fiction of the game?

Why don't countries like Japan just print more money?

Count All Possible Unique Combinations of Letters in a Word

Will generated tokens be progressively stronger when using Cathar's Crusade and Sorin, Grim Nemesis?

How can you guarantee that you won't change/quit job after just couple of months?

I found a password with hashcat, but it doesn't work

King or Queen-Which piece is which?

Is there any difference between Т34ВМ1 and КМ1858ВМ1/3?

Why are < or > requried to use /dev/tcp

Pythagorean trigonometry identity

How to prove this trigonometric expression?Proving an identity using reciprocal, quotient, or Pythagorean identities.Trigonometry Identity ProblemHarder Trigonometry Identity ($sec A+csc A$)Prove trigonometry identity for $sin A+cos A$Trig Identity / Pythagorean Theorem confusion?Trigonometry Identity (Proving)Trigonometry identity problemsProof of trignometric identityTrigonometry identity proof sumSecant and Tangent identity

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

3

$begingroup$

$$
fraccos(x)1-sin(x) + frac1-sin(x)cos(x) = 2sec(x)
$$

I have to prove that and I've been struggling for a very long time. Can anyone help me?

trigonometry

share|cite|improve this question

edited 8 hours ago

J. W. Tanner

8,3421723

asked 9 hours ago

ShafzKayShafzKay

211

New contributor

ShafzKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange. Please use MathJax
  $endgroup$
  – J. W. Tanner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Similar: math.stackexchange.com/q/385537/42969.
  $endgroup$
  – Martin R
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just cross-multiply
  $endgroup$
  – Ishan Deo
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As Ishan said, multiply both sides of the equation by $(1-sin(x))cdotcos(x)$.
  $endgroup$
  – irchans
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use $sin^2theta+cos^2theta=1$ at your leisure !
  $endgroup$
  – Antinous
  8 hours ago
  

  
  
  
  

 | 
show 1 more comment

3

$begingroup$

$$
fraccos(x)1-sin(x) + frac1-sin(x)cos(x) = 2sec(x)
$$

I have to prove that and I've been struggling for a very long time. Can anyone help me?

trigonometry

share|cite|improve this question

edited 8 hours ago

J. W. Tanner

8,3421723

asked 9 hours ago

ShafzKayShafzKay

211

New contributor

ShafzKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Mathematics Stack Exchange. Please use MathJax
  $endgroup$
  – J. W. Tanner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Similar: math.stackexchange.com/q/385537/42969.
  $endgroup$
  – Martin R
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Just cross-multiply
  $endgroup$
  – Ishan Deo
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As Ishan said, multiply both sides of the equation by $(1-sin(x))cdotcos(x)$.
  $endgroup$
  – irchans
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use $sin^2theta+cos^2theta=1$ at your leisure !
  $endgroup$
  – Antinous
  8 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

$$
fraccos(x)1-sin(x) + frac1-sin(x)cos(x) = 2sec(x)
$$

I have to prove that and I've been struggling for a very long time. Can anyone help me?

trigonometry

share|cite|improve this question

edited 8 hours ago

J. W. Tanner

8,3421723

asked 9 hours ago

ShafzKayShafzKay

211

New contributor

ShafzKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 8 hours ago

J. W. Tanner

8,3421723

asked 9 hours ago

ShafzKayShafzKay

211

New contributor

ShafzKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 8 hours ago

J. W. Tanner

8,3421723

asked 9 hours ago

ShafzKayShafzKay

211

New contributor

ShafzKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 8 hours ago

J. W. Tanner

8,3421723

edited 8 hours ago

J. W. Tanner

8,3421723

asked 9 hours ago

ShafzKayShafzKay

211

New contributor

ShafzKay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

ShafzKayShafzKay

211

5 Answers
5

active

oldest

votes

2

$begingroup$

We want to prove that $fraccos(x)1-sin(x)+frac1-sin(x)cos(x)$ is equal to $2sec(x)$

1. Multiply by $cos(x)$ and $(1-sin(x))$. You get:

$$fraccos^2(x)+(1-sin(x))^2(1-sin(x))cos(x)$$

2. By the Pythagorean Trig Identity $cos^2(x)=1-sin^2(x)$:

$$frac1-sin^2(x)+(1-sin(x))^2(1-sin(x))cos(x)$$

3. Expand the binomial in the numerator:

$$frac1-sin^2(x)+(1-2sin(x)+sin^2(x))(1-sin(x))cos(x)$$

4. Combine

$$frac2-2sin(x)(1-sin(x))cos(x)$$

5. Factor

$$frac2(1-sin(x))(1-sin(x))cos(x)$$

6. Simplify the fraction

$$frac2cos(x)$$

7. Use the Reciprocal Trig Identity $frac1cos(x) = sec(x)$
   

$$2*frac1cos(x) = 2sec(x)$$

$$therefore fraccos(x)1-sin(x)+frac1-sin(x)cos(x)=2sec(x)$$

Quick note: When you're trying to prove an identity like this, never manipulate both sides of the equations simultaneously (i.e. cross multiply). By doing so, you would be assuming that both sides are already equal before proving the identity.

share|cite|improve this answer

edited 7 hours ago

Théophile

20.9k13047

answered 8 hours ago

N. BarN. Bar

1987

$endgroup$

add a comment | 

2

$begingroup$

Other way
beginalign*
2sec x &= frac1cos x+frac1cos x
\
&=frac1cos x+frac1-sin x(1-sin x)cos x
\
&=frac1cos x+fracsin^2 x+cos^2 x-sin x(1-sin x)cos x
\
&=frac1cos x+frac-sin x(1-sin x)+cos^2 x(1-sin x)cos x
\
&=frac1cos x-fracsin x(1-sin x)(1-sin x)cos x+fraccos^2 x(1-sin x)cos x
\
&=frac1cos x-fracsin xcos x+fraccos x1-sin x
\
&=frac1-sin xcos x+fraccos x1-sin x
endalign*

share|cite|improve this answer

answered 7 hours ago

AsdrubalBeltranAsdrubalBeltran

3,21611018

$endgroup$

add a comment | 

1

$begingroup$

Hint

As $(1-sin x)(1+sin x)=cos^2x$

$dfraccos x1-sin x=dfrac1+sin x?$ for $cos xne0$

Alternatively use

https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution

share|cite|improve this answer

answered 8 hours ago

lab bhattacharjeelab bhattacharjee

233k15164285

$endgroup$

add a comment | 

1

$begingroup$

Multiply by $cos x$ and

$$
fraccos^2(x)+(1-sin(x))^21-sin(x)= frac1-2sin(x)+11-sin(x)= 2.
$$

share|cite|improve this answer

answered 7 hours ago

Yves DaoustYves Daoust

139k880237

$endgroup$

add a comment | 

0

$begingroup$

Since $dfraccos x1-sin x =dfraccos^2 xcos x(1-sin x)=dfrac1-sin^2 xcos x(1-sin x)=dfrac1+sin xcos x$,

$$
fraccos x1-sin x + frac1-sin xcos x =dfrac2cos x=2sec x
$$

share|cite|improve this answer

answered 7 hours ago

CY AriesCY Aries

21.2k12049

$endgroup$

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
);



);

ShafzKay is a new contributor. Be nice, and check out our Code of Conduct.

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%2f3266456%2fpythagorean-trigonometry-identity%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

We want to prove that $fraccos(x)1-sin(x)+frac1-sin(x)cos(x)$ is equal to $2sec(x)$

1. Multiply by $cos(x)$ and $(1-sin(x))$. You get:

$$fraccos^2(x)+(1-sin(x))^2(1-sin(x))cos(x)$$

2. By the Pythagorean Trig Identity $cos^2(x)=1-sin^2(x)$:

$$frac1-sin^2(x)+(1-sin(x))^2(1-sin(x))cos(x)$$

3. Expand the binomial in the numerator:

$$frac1-sin^2(x)+(1-2sin(x)+sin^2(x))(1-sin(x))cos(x)$$

4. Combine

$$frac2-2sin(x)(1-sin(x))cos(x)$$

5. Factor

$$frac2(1-sin(x))(1-sin(x))cos(x)$$

6. Simplify the fraction

$$frac2cos(x)$$

7. Use the Reciprocal Trig Identity $frac1cos(x) = sec(x)$
   

$$2*frac1cos(x) = 2sec(x)$$

$$therefore fraccos(x)1-sin(x)+frac1-sin(x)cos(x)=2sec(x)$$

Quick note: When you're trying to prove an identity like this, never manipulate both sides of the equations simultaneously (i.e. cross multiply). By doing so, you would be assuming that both sides are already equal before proving the identity.

share|cite|improve this answer

edited 7 hours ago

Théophile

20.9k13047

answered 8 hours ago

N. BarN. Bar

1987

$endgroup$

add a comment | 

2

$begingroup$

We want to prove that $fraccos(x)1-sin(x)+frac1-sin(x)cos(x)$ is equal to $2sec(x)$

1. Multiply by $cos(x)$ and $(1-sin(x))$. You get:

$$fraccos^2(x)+(1-sin(x))^2(1-sin(x))cos(x)$$

2. By the Pythagorean Trig Identity $cos^2(x)=1-sin^2(x)$:

$$frac1-sin^2(x)+(1-sin(x))^2(1-sin(x))cos(x)$$

3. Expand the binomial in the numerator:

$$frac1-sin^2(x)+(1-2sin(x)+sin^2(x))(1-sin(x))cos(x)$$

4. Combine

$$frac2-2sin(x)(1-sin(x))cos(x)$$

5. Factor

$$frac2(1-sin(x))(1-sin(x))cos(x)$$

6. Simplify the fraction

$$frac2cos(x)$$

7. Use the Reciprocal Trig Identity $frac1cos(x) = sec(x)$
   

$$2*frac1cos(x) = 2sec(x)$$

$$therefore fraccos(x)1-sin(x)+frac1-sin(x)cos(x)=2sec(x)$$

Quick note: When you're trying to prove an identity like this, never manipulate both sides of the equations simultaneously (i.e. cross multiply). By doing so, you would be assuming that both sides are already equal before proving the identity.

share|cite|improve this answer

edited 7 hours ago

Théophile

20.9k13047

answered 8 hours ago

N. BarN. Bar

1987

$endgroup$

add a comment | 

$begingroup$

We want to prove that $fraccos(x)1-sin(x)+frac1-sin(x)cos(x)$ is equal to $2sec(x)$

1. Multiply by $cos(x)$ and $(1-sin(x))$. You get:

$$fraccos^2(x)+(1-sin(x))^2(1-sin(x))cos(x)$$

2. By the Pythagorean Trig Identity $cos^2(x)=1-sin^2(x)$:

$$frac1-sin^2(x)+(1-sin(x))^2(1-sin(x))cos(x)$$

3. Expand the binomial in the numerator:

$$frac1-sin^2(x)+(1-2sin(x)+sin^2(x))(1-sin(x))cos(x)$$

4. Combine

$$frac2-2sin(x)(1-sin(x))cos(x)$$

5. Factor

$$frac2(1-sin(x))(1-sin(x))cos(x)$$

6. Simplify the fraction

$$frac2cos(x)$$

7. Use the Reciprocal Trig Identity $frac1cos(x) = sec(x)$
   

$$2*frac1cos(x) = 2sec(x)$$

$$therefore fraccos(x)1-sin(x)+frac1-sin(x)cos(x)=2sec(x)$$

Quick note: When you're trying to prove an identity like this, never manipulate both sides of the equations simultaneously (i.e. cross multiply). By doing so, you would be assuming that both sides are already equal before proving the identity.

share|cite|improve this answer

edited 7 hours ago

Théophile

20.9k13047

answered 8 hours ago

N. BarN. Bar

1987

$endgroup$

share|cite|improve this answer

edited 7 hours ago

Théophile

20.9k13047

answered 8 hours ago

N. BarN. Bar

1987

edited 7 hours ago

Théophile

20.9k13047

edited 7 hours ago

Théophile

20.9k13047

answered 8 hours ago

N. BarN. Bar

1987

answered 8 hours ago

N. BarN. Bar

1987

2

$begingroup$

Other way
beginalign*
2sec x &= frac1cos x+frac1cos x
\
&=frac1cos x+frac1-sin x(1-sin x)cos x
\
&=frac1cos x+fracsin^2 x+cos^2 x-sin x(1-sin x)cos x
\
&=frac1cos x+frac-sin x(1-sin x)+cos^2 x(1-sin x)cos x
\
&=frac1cos x-fracsin x(1-sin x)(1-sin x)cos x+fraccos^2 x(1-sin x)cos x
\
&=frac1cos x-fracsin xcos x+fraccos x1-sin x
\
&=frac1-sin xcos x+fraccos x1-sin x
endalign*

share|cite|improve this answer

answered 7 hours ago

AsdrubalBeltranAsdrubalBeltran

3,21611018

$endgroup$

add a comment | 

2

$begingroup$

Other way
beginalign*
2sec x &= frac1cos x+frac1cos x
\
&=frac1cos x+frac1-sin x(1-sin x)cos x
\
&=frac1cos x+fracsin^2 x+cos^2 x-sin x(1-sin x)cos x
\
&=frac1cos x+frac-sin x(1-sin x)+cos^2 x(1-sin x)cos x
\
&=frac1cos x-fracsin x(1-sin x)(1-sin x)cos x+fraccos^2 x(1-sin x)cos x
\
&=frac1cos x-fracsin xcos x+fraccos x1-sin x
\
&=frac1-sin xcos x+fraccos x1-sin x
endalign*

share|cite|improve this answer

answered 7 hours ago

AsdrubalBeltranAsdrubalBeltran

3,21611018

$endgroup$

add a comment | 

$begingroup$

Other way
beginalign*
2sec x &= frac1cos x+frac1cos x
\
&=frac1cos x+frac1-sin x(1-sin x)cos x
\
&=frac1cos x+fracsin^2 x+cos^2 x-sin x(1-sin x)cos x
\
&=frac1cos x+frac-sin x(1-sin x)+cos^2 x(1-sin x)cos x
\
&=frac1cos x-fracsin x(1-sin x)(1-sin x)cos x+fraccos^2 x(1-sin x)cos x
\
&=frac1cos x-fracsin xcos x+fraccos x1-sin x
\
&=frac1-sin xcos x+fraccos x1-sin x
endalign*

share|cite|improve this answer

answered 7 hours ago

AsdrubalBeltranAsdrubalBeltran

3,21611018

$endgroup$

share|cite|improve this answer

answered 7 hours ago

AsdrubalBeltranAsdrubalBeltran

3,21611018

answered 7 hours ago

AsdrubalBeltranAsdrubalBeltran

3,21611018

answered 7 hours ago

AsdrubalBeltranAsdrubalBeltran

3,21611018

1

$begingroup$

Hint

As $(1-sin x)(1+sin x)=cos^2x$

$dfraccos x1-sin x=dfrac1+sin x?$ for $cos xne0$

Alternatively use

https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution

share|cite|improve this answer

answered 8 hours ago

lab bhattacharjeelab bhattacharjee

233k15164285

$endgroup$

add a comment | 

1

$begingroup$

Hint

As $(1-sin x)(1+sin x)=cos^2x$

$dfraccos x1-sin x=dfrac1+sin x?$ for $cos xne0$

Alternatively use

https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution

share|cite|improve this answer

answered 8 hours ago

lab bhattacharjeelab bhattacharjee

233k15164285

$endgroup$

add a comment | 

$begingroup$

Hint

As $(1-sin x)(1+sin x)=cos^2x$

$dfraccos x1-sin x=dfrac1+sin x?$ for $cos xne0$

Alternatively use

https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution

share|cite|improve this answer

answered 8 hours ago

lab bhattacharjeelab bhattacharjee

233k15164285

$endgroup$

share|cite|improve this answer

answered 8 hours ago

lab bhattacharjeelab bhattacharjee

233k15164285

answered 8 hours ago

lab bhattacharjeelab bhattacharjee

233k15164285

answered 8 hours ago

lab bhattacharjeelab bhattacharjee

233k15164285

1

$begingroup$

Multiply by $cos x$ and

$$
fraccos^2(x)+(1-sin(x))^21-sin(x)= frac1-2sin(x)+11-sin(x)= 2.
$$

share|cite|improve this answer

answered 7 hours ago

Yves DaoustYves Daoust

139k880237

$endgroup$

add a comment | 

1

$begingroup$

Multiply by $cos x$ and

$$
fraccos^2(x)+(1-sin(x))^21-sin(x)= frac1-2sin(x)+11-sin(x)= 2.
$$

share|cite|improve this answer

answered 7 hours ago

Yves DaoustYves Daoust

139k880237

$endgroup$

add a comment | 

$begingroup$

Multiply by $cos x$ and

$$
fraccos^2(x)+(1-sin(x))^21-sin(x)= frac1-2sin(x)+11-sin(x)= 2.
$$

share|cite|improve this answer

answered 7 hours ago

Yves DaoustYves Daoust

139k880237

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Yves DaoustYves Daoust

139k880237

answered 7 hours ago

Yves DaoustYves Daoust

139k880237

answered 7 hours ago

Yves DaoustYves Daoust

139k880237

0

$begingroup$

Since $dfraccos x1-sin x =dfraccos^2 xcos x(1-sin x)=dfrac1-sin^2 xcos x(1-sin x)=dfrac1+sin xcos x$,

$$
fraccos x1-sin x + frac1-sin xcos x =dfrac2cos x=2sec x
$$

share|cite|improve this answer

answered 7 hours ago

CY AriesCY Aries

21.2k12049

$endgroup$

add a comment | 

0

$begingroup$

Since $dfraccos x1-sin x =dfraccos^2 xcos x(1-sin x)=dfrac1-sin^2 xcos x(1-sin x)=dfrac1+sin xcos x$,

$$
fraccos x1-sin x + frac1-sin xcos x =dfrac2cos x=2sec x
$$

share|cite|improve this answer

answered 7 hours ago

CY AriesCY Aries

21.2k12049

$endgroup$

add a comment | 

$begingroup$

Since $dfraccos x1-sin x =dfraccos^2 xcos x(1-sin x)=dfrac1-sin^2 xcos x(1-sin x)=dfrac1+sin xcos x$,

$$
fraccos x1-sin x + frac1-sin xcos x =dfrac2cos x=2sec x
$$

share|cite|improve this answer

answered 7 hours ago

CY AriesCY Aries

21.2k12049

$endgroup$

share|cite|improve this answer

answered 7 hours ago

CY AriesCY Aries

21.2k12049

answered 7 hours ago

CY AriesCY Aries

21.2k12049

answered 7 hours ago

CY AriesCY Aries

21.2k12049