Proving $fracpsin x+ncos x=fracpcsc(arctan(n)+x)sin(arctan(n)) n$

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The following two expressions produce the same graph.
$$fracpsin(x)+ncos(x) = frac pcsc(arctan(n)+x)sin(arctan(n))n.$$
How to prove that they are equal?

If drawn with polar coordinates, $x$ is $theta$, the $p$ is the the point of intersection with the $y$-axis and $-n$ is the slope.

trigonometry polar-coordinates

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edited Jul 29 at 1:40

Math Lover

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up vote
2
down vote

favorite

The following two expressions produce the same graph.
$$fracpsin(x)+ncos(x) = frac pcsc(arctan(n)+x)sin(arctan(n))n.$$
How to prove that they are equal?

If drawn with polar coordinates, $x$ is $theta$, the $p$ is the the point of intersection with the $y$-axis and $-n$ is the slope.

trigonometry polar-coordinates

share|cite|improve this question

edited Jul 29 at 1:40

Math Lover

12.3k21232

asked Jul 29 at 0:33

user37421

489411

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

The following two expressions produce the same graph.
$$fracpsin(x)+ncos(x) = frac pcsc(arctan(n)+x)sin(arctan(n))n.$$
How to prove that they are equal?

If drawn with polar coordinates, $x$ is $theta$, the $p$ is the the point of intersection with the $y$-axis and $-n$ is the slope.

trigonometry polar-coordinates

share|cite|improve this question

edited Jul 29 at 1:40

Math Lover

12.3k21232

asked Jul 29 at 0:33

user37421

489411

The following two expressions produce the same graph.
$$fracpsin(x)+ncos(x) = frac pcsc(arctan(n)+x)sin(arctan(n))n.$$
How to prove that they are equal?

If drawn with polar coordinates, $x$ is $theta$, the $p$ is the the point of intersection with the $y$-axis and $-n$ is the slope.

trigonometry polar-coordinates

share|cite|improve this question

edited Jul 29 at 1:40

Math Lover

12.3k21232

asked Jul 29 at 0:33

user37421

489411

share|cite|improve this question

share|cite|improve this question

edited Jul 29 at 1:40

Math Lover

12.3k21232

edited Jul 29 at 1:40

Math Lover

12.3k21232

edited Jul 29 at 1:40

Math Lover

12.3k21232

12.3k21232

asked Jul 29 at 0:33

user37421

489411

asked Jul 29 at 0:33

user37421

489411

asked Jul 29 at 0:33

user37421

489411

489411

add a comment | 

add a comment | 

3 Answers
3

active

oldest

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up vote
1
down vote



accepted

Hint:

$$arctan(n) = y implies n = tan(y) implies sin(y) = fracnsqrtn^2+1, textand cos(y) = frac1sqrtn^2+1.$$
Also,
$$csc(arctan(n)+x) = frac1sin(arctan(n)+x) = frac1sin(arctan(n))cos(x)+cos(arctan(n))sin(x)$$

share|cite|improve this answer

answered Jul 29 at 1:37

Math Lover

12.3k21232

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up vote
1
down vote

Whenever you are dealing with expressions like $sin(arctan(n))$, the trick is to draw a triangle like below:

You can construct this by noting that for the angle to be $arctan(n)$ the opposite and adjacent sides must be $n$ and $1$ respecitvely, in which case the Pythagoras rule gives the hypotenuse.

From this you can see:
$$sin(arctan(n))=fracnsqrt1+n^2,$$
$$cos(arctan(n))=frac1sqrt1+n^2.$$
You will also want to expand $sin(arctan(n)+x)$, which you can do using the identity $sin(a+b)=sin(a)cos(b)+cos(a)sin(b)$.

share|cite|improve this answer

edited Jul 29 at 1:42

answered Jul 29 at 1:36

Ruvi Lecamwasam

1,230618

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's not $cos(x)$; it's $csc(x)$.
  – Michael McGovern
  Jul 29 at 1:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I'll fix that
  – Ruvi Lecamwasam
  Jul 29 at 1:42
  

  
  
  
  

add a comment | 

up vote
1
down vote

Hint:

$$sin(alpha+beta)=sin(alpha)cos(beta)+cos(alpha)sin(beta)$$

and

$$sin(arctan(x))=fracxsqrt1+x^2$$

share|cite|improve this answer

edited Jul 29 at 1:44

Math Lover

12.3k21232

answered Jul 29 at 1:33

Michael McGovern

5701314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just letting you know, your two equations here are showing on the same line.
  – Ruvi Lecamwasam
  Jul 29 at 1:41
  

  
  
  
  

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Hint:

$$arctan(n) = y implies n = tan(y) implies sin(y) = fracnsqrtn^2+1, textand cos(y) = frac1sqrtn^2+1.$$
Also,
$$csc(arctan(n)+x) = frac1sin(arctan(n)+x) = frac1sin(arctan(n))cos(x)+cos(arctan(n))sin(x)$$

share|cite|improve this answer

answered Jul 29 at 1:37

Math Lover

12.3k21232

add a comment | 

up vote
1
down vote



accepted

Hint:

$$arctan(n) = y implies n = tan(y) implies sin(y) = fracnsqrtn^2+1, textand cos(y) = frac1sqrtn^2+1.$$
Also,
$$csc(arctan(n)+x) = frac1sin(arctan(n)+x) = frac1sin(arctan(n))cos(x)+cos(arctan(n))sin(x)$$

share|cite|improve this answer

answered Jul 29 at 1:37

Math Lover

12.3k21232

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Hint:

$$arctan(n) = y implies n = tan(y) implies sin(y) = fracnsqrtn^2+1, textand cos(y) = frac1sqrtn^2+1.$$
Also,
$$csc(arctan(n)+x) = frac1sin(arctan(n)+x) = frac1sin(arctan(n))cos(x)+cos(arctan(n))sin(x)$$

share|cite|improve this answer

answered Jul 29 at 1:37

Math Lover

12.3k21232

Hint:

$$arctan(n) = y implies n = tan(y) implies sin(y) = fracnsqrtn^2+1, textand cos(y) = frac1sqrtn^2+1.$$
Also,
$$csc(arctan(n)+x) = frac1sin(arctan(n)+x) = frac1sin(arctan(n))cos(x)+cos(arctan(n))sin(x)$$

share|cite|improve this answer

answered Jul 29 at 1:37

Math Lover

12.3k21232

share|cite|improve this answer

share|cite|improve this answer

answered Jul 29 at 1:37

Math Lover

12.3k21232

answered Jul 29 at 1:37

Math Lover

12.3k21232

answered Jul 29 at 1:37

Math Lover

12.3k21232

12.3k21232

add a comment | 

add a comment | 

up vote
1
down vote

Whenever you are dealing with expressions like $sin(arctan(n))$, the trick is to draw a triangle like below:

You can construct this by noting that for the angle to be $arctan(n)$ the opposite and adjacent sides must be $n$ and $1$ respecitvely, in which case the Pythagoras rule gives the hypotenuse.

From this you can see:
$$sin(arctan(n))=fracnsqrt1+n^2,$$
$$cos(arctan(n))=frac1sqrt1+n^2.$$
You will also want to expand $sin(arctan(n)+x)$, which you can do using the identity $sin(a+b)=sin(a)cos(b)+cos(a)sin(b)$.

share|cite|improve this answer

edited Jul 29 at 1:42

answered Jul 29 at 1:36

Ruvi Lecamwasam

1,230618

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's not $cos(x)$; it's $csc(x)$.
  – Michael McGovern
  Jul 29 at 1:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I'll fix that
  – Ruvi Lecamwasam
  Jul 29 at 1:42
  

  
  
  
  

add a comment | 

up vote
1
down vote

Whenever you are dealing with expressions like $sin(arctan(n))$, the trick is to draw a triangle like below:

You can construct this by noting that for the angle to be $arctan(n)$ the opposite and adjacent sides must be $n$ and $1$ respecitvely, in which case the Pythagoras rule gives the hypotenuse.

From this you can see:
$$sin(arctan(n))=fracnsqrt1+n^2,$$
$$cos(arctan(n))=frac1sqrt1+n^2.$$
You will also want to expand $sin(arctan(n)+x)$, which you can do using the identity $sin(a+b)=sin(a)cos(b)+cos(a)sin(b)$.

share|cite|improve this answer

edited Jul 29 at 1:42

answered Jul 29 at 1:36

Ruvi Lecamwasam

1,230618

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's not $cos(x)$; it's $csc(x)$.
  – Michael McGovern
  Jul 29 at 1:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I'll fix that
  – Ruvi Lecamwasam
  Jul 29 at 1:42
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Whenever you are dealing with expressions like $sin(arctan(n))$, the trick is to draw a triangle like below:

You can construct this by noting that for the angle to be $arctan(n)$ the opposite and adjacent sides must be $n$ and $1$ respecitvely, in which case the Pythagoras rule gives the hypotenuse.

From this you can see:
$$sin(arctan(n))=fracnsqrt1+n^2,$$
$$cos(arctan(n))=frac1sqrt1+n^2.$$
You will also want to expand $sin(arctan(n)+x)$, which you can do using the identity $sin(a+b)=sin(a)cos(b)+cos(a)sin(b)$.

share|cite|improve this answer

edited Jul 29 at 1:42

answered Jul 29 at 1:36

Ruvi Lecamwasam

1,230618

Whenever you are dealing with expressions like $sin(arctan(n))$, the trick is to draw a triangle like below:

You can construct this by noting that for the angle to be $arctan(n)$ the opposite and adjacent sides must be $n$ and $1$ respecitvely, in which case the Pythagoras rule gives the hypotenuse.

From this you can see:
$$sin(arctan(n))=fracnsqrt1+n^2,$$
$$cos(arctan(n))=frac1sqrt1+n^2.$$
You will also want to expand $sin(arctan(n)+x)$, which you can do using the identity $sin(a+b)=sin(a)cos(b)+cos(a)sin(b)$.

share|cite|improve this answer

edited Jul 29 at 1:42

answered Jul 29 at 1:36

Ruvi Lecamwasam

1,230618

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 1:42

edited Jul 29 at 1:42

edited Jul 29 at 1:42

answered Jul 29 at 1:36

Ruvi Lecamwasam

1,230618

answered Jul 29 at 1:36

Ruvi Lecamwasam

1,230618

answered Jul 29 at 1:36

Ruvi Lecamwasam

1,230618

1,230618

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's not $cos(x)$; it's $csc(x)$.
  – Michael McGovern
  Jul 29 at 1:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I'll fix that
  – Ruvi Lecamwasam
  Jul 29 at 1:42
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It's not $cos(x)$; it's $csc(x)$.
  – Michael McGovern
  Jul 29 at 1:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks, I'll fix that
  – Ruvi Lecamwasam
  Jul 29 at 1:42
  

  
  
  
  

It's not $cos(x)$; it's $csc(x)$.
– Michael McGovern
Jul 29 at 1:38

It's not $cos(x)$; it's $csc(x)$.
– Michael McGovern
Jul 29 at 1:38

Thanks, I'll fix that
– Ruvi Lecamwasam
Jul 29 at 1:42

Thanks, I'll fix that
– Ruvi Lecamwasam
Jul 29 at 1:42

add a comment | 

up vote
1
down vote

Hint:

$$sin(alpha+beta)=sin(alpha)cos(beta)+cos(alpha)sin(beta)$$

and

$$sin(arctan(x))=fracxsqrt1+x^2$$

share|cite|improve this answer

edited Jul 29 at 1:44

Math Lover

12.3k21232

answered Jul 29 at 1:33

Michael McGovern

5701314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just letting you know, your two equations here are showing on the same line.
  – Ruvi Lecamwasam
  Jul 29 at 1:41
  

  
  
  
  

add a comment | 

up vote
1
down vote

Hint:

$$sin(alpha+beta)=sin(alpha)cos(beta)+cos(alpha)sin(beta)$$

and

$$sin(arctan(x))=fracxsqrt1+x^2$$

share|cite|improve this answer

edited Jul 29 at 1:44

Math Lover

12.3k21232

answered Jul 29 at 1:33

Michael McGovern

5701314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just letting you know, your two equations here are showing on the same line.
  – Ruvi Lecamwasam
  Jul 29 at 1:41
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint:

$$sin(alpha+beta)=sin(alpha)cos(beta)+cos(alpha)sin(beta)$$

and

$$sin(arctan(x))=fracxsqrt1+x^2$$

share|cite|improve this answer

edited Jul 29 at 1:44

Math Lover

12.3k21232

answered Jul 29 at 1:33

Michael McGovern

5701314

Hint:

$$sin(alpha+beta)=sin(alpha)cos(beta)+cos(alpha)sin(beta)$$

and

$$sin(arctan(x))=fracxsqrt1+x^2$$

share|cite|improve this answer

edited Jul 29 at 1:44

Math Lover

12.3k21232

answered Jul 29 at 1:33

Michael McGovern

5701314

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 1:44

Math Lover

12.3k21232

edited Jul 29 at 1:44

Math Lover

12.3k21232

edited Jul 29 at 1:44

Math Lover

12.3k21232

12.3k21232

answered Jul 29 at 1:33

Michael McGovern

5701314

answered Jul 29 at 1:33

Michael McGovern

5701314

answered Jul 29 at 1:33

Michael McGovern

5701314

5701314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just letting you know, your two equations here are showing on the same line.
  – Ruvi Lecamwasam
  Jul 29 at 1:41
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just letting you know, your two equations here are showing on the same line.
  – Ruvi Lecamwasam
  Jul 29 at 1:41
  

  
  
  
  

Just letting you know, your two equations here are showing on the same line.
– Ruvi Lecamwasam
Jul 29 at 1:41

Just letting you know, your two equations here are showing on the same line.
– Ruvi Lecamwasam
Jul 29 at 1:41

add a comment | 

 

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