Proving a long-ish trig identity

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Prove this identity without using cross multiplication by manipulating one side using trig identities:

$$fracsin^3x-cos^3xsin x+cos x = fraccsc^2x-cot x-2cos^2x1-cot^2x$$

I first started off on the LHS and managed to get the denominator to become $1-cot^2x$ by multiplying by $sin x - cos x$ and then dividing by $sin^2 x$, but from there I had no idea how to continue.

algebra-precalculus trigonometry

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edited Jul 23 at 20:34

Bernard

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asked Jul 23 at 20:28

Jessca

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  Start with the RHS, convert everything to sin and cos, it is easier to work that way. Then simplify.
  – Isko10986
  Jul 23 at 20:31
  

  
  
  
  

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Prove this identity without using cross multiplication by manipulating one side using trig identities:

$$fracsin^3x-cos^3xsin x+cos x = fraccsc^2x-cot x-2cos^2x1-cot^2x$$

I first started off on the LHS and managed to get the denominator to become $1-cot^2x$ by multiplying by $sin x - cos x$ and then dividing by $sin^2 x$, but from there I had no idea how to continue.

algebra-precalculus trigonometry

share|cite|improve this question

edited Jul 23 at 20:34

Bernard

110k635103

asked Jul 23 at 20:28

Jessca

32

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Start with the RHS, convert everything to sin and cos, it is easier to work that way. Then simplify.
  – Isko10986
  Jul 23 at 20:31
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Prove this identity without using cross multiplication by manipulating one side using trig identities:

$$fracsin^3x-cos^3xsin x+cos x = fraccsc^2x-cot x-2cos^2x1-cot^2x$$

I first started off on the LHS and managed to get the denominator to become $1-cot^2x$ by multiplying by $sin x - cos x$ and then dividing by $sin^2 x$, but from there I had no idea how to continue.

algebra-precalculus trigonometry

share|cite|improve this question

edited Jul 23 at 20:34

Bernard

110k635103

asked Jul 23 at 20:28

Jessca

32

Prove this identity without using cross multiplication by manipulating one side using trig identities:

$$fracsin^3x-cos^3xsin x+cos x = fraccsc^2x-cot x-2cos^2x1-cot^2x$$

I first started off on the LHS and managed to get the denominator to become $1-cot^2x$ by multiplying by $sin x - cos x$ and then dividing by $sin^2 x$, but from there I had no idea how to continue.

algebra-precalculus trigonometry

share|cite|improve this question

edited Jul 23 at 20:34

Bernard

110k635103

asked Jul 23 at 20:28

Jessca

32

share|cite|improve this question

share|cite|improve this question

edited Jul 23 at 20:34

Bernard

110k635103

edited Jul 23 at 20:34

Bernard

110k635103

edited Jul 23 at 20:34

Bernard

110k635103

110k635103

asked Jul 23 at 20:28

Jessca

32

asked Jul 23 at 20:28

Jessca

32

asked Jul 23 at 20:28

Jessca

32

32

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Start with the RHS, convert everything to sin and cos, it is easier to work that way. Then simplify.
  – Isko10986
  Jul 23 at 20:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Start with the RHS, convert everything to sin and cos, it is easier to work that way. Then simplify.
  – Isko10986
  Jul 23 at 20:31
  

  
  
  
  

2

2

Start with the RHS, convert everything to sin and cos, it is easier to work that way. Then simplify.
– Isko10986
Jul 23 at 20:31

Start with the RHS, convert everything to sin and cos, it is easier to work that way. Then simplify.
– Isko10986
Jul 23 at 20:31

add a comment | 

4 Answers
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accepted

Let $s= sin x $ and $c=cos x$
begineqnarray*
fracs^3-c^3s+c fracs-cs-c = fracs^4+c^4-sc(s^2+c^2)s^2-c^2.
endeqnarray*
Now $s^4+c^4=(s^2+c^2)^2-2s^2 c^2$ and divide top & bottom by $s^2$
begineqnarray*
frac1/s^2-c/s-2c^21-c^2/s^2 .
endeqnarray*

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answered Jul 23 at 20:38

Donald Splutterwit

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up vote
1
down vote

Note that
beginalign
sin^3(x)-cos^3(x) & = (sin(x)-cos(x))(sin^2(x)+cos^2(x)+sin(x)cos(x)) \
& =(sin(x)-cos(x))(1+sin(x)cos(x)).
endalign
Therefore,
beginalign
fracsin^3(x)-cos^3(x)sin(x)+cos(x) & = frac(sin(x)-cos(x))(1+sin(x)cos(x))sin(x)+cos(x) \
& = frac(sin(x)-cos(x))^2(1+sin(x)cos(x))sin^2(x)-cos^2(x) \
& = frac(1-2sin(x)cos(x))(1+sin(x)cos(x))sin^2(x)-cos^2(x) \
& = frac1-sin(x)cos(x)-2sin^2(x)cos^2(x)sin^2(x)-cos^2(x) \
& = fracsin^2(x)sin^2(x)fraccsc^2(x)-cot(x)-2cos^2(x)1-cot^2(x).
endalign

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answered Jul 23 at 20:44

Math Lover

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up vote
1
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I'd start with the right-hand side:
beginalign
fraccsc^2x-cot x-2cos^2x1-cot^2x
&=fracdfrac1sin^2x-dfraccos xsin x-2cos^2x1-dfraccos^2xsin^2x
\[6px]
&=frac1-sin xcos x-2cos^2xsin^2xsin^2x-cos^2x
\[6px]
&=frac(sin^2x+cos^2x)^2-(sin^2x+cos^2x)sin xcos x-2sin^2xcos^2xsin^2x-cos^2x
\[6px]
&=fracsin^4x-sin^3xcos x-sin xcos^3x+cos^4xsin^2x-cos^2x
\[6px]
&=fracsin^3x(sin x-cos x)-cos^3x(sin x-cos x)sin^2x-cos^2x
\[6px]
&=frac(sin^3x-cos^3x)(sin x-cos x)(sin x+cos x)(sin x-cos x)
\[6px]
&=fracsin^3x-cos^3xsin x+cos x
endalign
The trick is to “make the numerator homogeneous”.

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answered Jul 23 at 21:01

egreg

164k1180187

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Use that $$sin^2(x)(1-cot(x)^2)=sin(x)^2-cos(x)^2$$
then we get

$$fracsin(x)^3-cos(x)^3sin(x)+cos(x)=frac2cos(x)^4-2cos(x)^2-sin(x)cos(x)+1(sin(x)+cos(x))(sin(x)-cos(x))$$
and then use that
$$sin(x)^3-cos(x)^3=(sin(x)-cos(x))(sin(x)^2+sin(x)cos(x)+cos(x)^2)$$

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edited Jul 23 at 20:46

answered Jul 23 at 20:40

Dr. Sonnhard Graubner

66.7k32659

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Let $s= sin x $ and $c=cos x$
begineqnarray*
fracs^3-c^3s+c fracs-cs-c = fracs^4+c^4-sc(s^2+c^2)s^2-c^2.
endeqnarray*
Now $s^4+c^4=(s^2+c^2)^2-2s^2 c^2$ and divide top & bottom by $s^2$
begineqnarray*
frac1/s^2-c/s-2c^21-c^2/s^2 .
endeqnarray*

share|cite|improve this answer

answered Jul 23 at 20:38

Donald Splutterwit

21.3k21243

add a comment | 

up vote
2
down vote



accepted

Let $s= sin x $ and $c=cos x$
begineqnarray*
fracs^3-c^3s+c fracs-cs-c = fracs^4+c^4-sc(s^2+c^2)s^2-c^2.
endeqnarray*
Now $s^4+c^4=(s^2+c^2)^2-2s^2 c^2$ and divide top & bottom by $s^2$
begineqnarray*
frac1/s^2-c/s-2c^21-c^2/s^2 .
endeqnarray*

share|cite|improve this answer

answered Jul 23 at 20:38

Donald Splutterwit

21.3k21243

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Let $s= sin x $ and $c=cos x$
begineqnarray*
fracs^3-c^3s+c fracs-cs-c = fracs^4+c^4-sc(s^2+c^2)s^2-c^2.
endeqnarray*
Now $s^4+c^4=(s^2+c^2)^2-2s^2 c^2$ and divide top & bottom by $s^2$
begineqnarray*
frac1/s^2-c/s-2c^21-c^2/s^2 .
endeqnarray*

share|cite|improve this answer

answered Jul 23 at 20:38

Donald Splutterwit

21.3k21243

Let $s= sin x $ and $c=cos x$
begineqnarray*
fracs^3-c^3s+c fracs-cs-c = fracs^4+c^4-sc(s^2+c^2)s^2-c^2.
endeqnarray*
Now $s^4+c^4=(s^2+c^2)^2-2s^2 c^2$ and divide top & bottom by $s^2$
begineqnarray*
frac1/s^2-c/s-2c^21-c^2/s^2 .
endeqnarray*

share|cite|improve this answer

answered Jul 23 at 20:38

Donald Splutterwit

21.3k21243

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share|cite|improve this answer

answered Jul 23 at 20:38

Donald Splutterwit

21.3k21243

answered Jul 23 at 20:38

Donald Splutterwit

21.3k21243

answered Jul 23 at 20:38

Donald Splutterwit

21.3k21243

21.3k21243

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add a comment | 

up vote
1
down vote

Note that
beginalign
sin^3(x)-cos^3(x) & = (sin(x)-cos(x))(sin^2(x)+cos^2(x)+sin(x)cos(x)) \
& =(sin(x)-cos(x))(1+sin(x)cos(x)).
endalign
Therefore,
beginalign
fracsin^3(x)-cos^3(x)sin(x)+cos(x) & = frac(sin(x)-cos(x))(1+sin(x)cos(x))sin(x)+cos(x) \
& = frac(sin(x)-cos(x))^2(1+sin(x)cos(x))sin^2(x)-cos^2(x) \
& = frac(1-2sin(x)cos(x))(1+sin(x)cos(x))sin^2(x)-cos^2(x) \
& = frac1-sin(x)cos(x)-2sin^2(x)cos^2(x)sin^2(x)-cos^2(x) \
& = fracsin^2(x)sin^2(x)fraccsc^2(x)-cot(x)-2cos^2(x)1-cot^2(x).
endalign

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answered Jul 23 at 20:44

Math Lover

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up vote
1
down vote

Note that
beginalign
sin^3(x)-cos^3(x) & = (sin(x)-cos(x))(sin^2(x)+cos^2(x)+sin(x)cos(x)) \
& =(sin(x)-cos(x))(1+sin(x)cos(x)).
endalign
Therefore,
beginalign
fracsin^3(x)-cos^3(x)sin(x)+cos(x) & = frac(sin(x)-cos(x))(1+sin(x)cos(x))sin(x)+cos(x) \
& = frac(sin(x)-cos(x))^2(1+sin(x)cos(x))sin^2(x)-cos^2(x) \
& = frac(1-2sin(x)cos(x))(1+sin(x)cos(x))sin^2(x)-cos^2(x) \
& = frac1-sin(x)cos(x)-2sin^2(x)cos^2(x)sin^2(x)-cos^2(x) \
& = fracsin^2(x)sin^2(x)fraccsc^2(x)-cot(x)-2cos^2(x)1-cot^2(x).
endalign

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answered Jul 23 at 20:44

Math Lover

12.3k21232

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up vote
1
down vote

up vote
1
down vote

Note that
beginalign
sin^3(x)-cos^3(x) & = (sin(x)-cos(x))(sin^2(x)+cos^2(x)+sin(x)cos(x)) \
& =(sin(x)-cos(x))(1+sin(x)cos(x)).
endalign
Therefore,
beginalign
fracsin^3(x)-cos^3(x)sin(x)+cos(x) & = frac(sin(x)-cos(x))(1+sin(x)cos(x))sin(x)+cos(x) \
& = frac(sin(x)-cos(x))^2(1+sin(x)cos(x))sin^2(x)-cos^2(x) \
& = frac(1-2sin(x)cos(x))(1+sin(x)cos(x))sin^2(x)-cos^2(x) \
& = frac1-sin(x)cos(x)-2sin^2(x)cos^2(x)sin^2(x)-cos^2(x) \
& = fracsin^2(x)sin^2(x)fraccsc^2(x)-cot(x)-2cos^2(x)1-cot^2(x).
endalign

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answered Jul 23 at 20:44

Math Lover

12.3k21232

Note that
beginalign
sin^3(x)-cos^3(x) & = (sin(x)-cos(x))(sin^2(x)+cos^2(x)+sin(x)cos(x)) \
& =(sin(x)-cos(x))(1+sin(x)cos(x)).
endalign
Therefore,
beginalign
fracsin^3(x)-cos^3(x)sin(x)+cos(x) & = frac(sin(x)-cos(x))(1+sin(x)cos(x))sin(x)+cos(x) \
& = frac(sin(x)-cos(x))^2(1+sin(x)cos(x))sin^2(x)-cos^2(x) \
& = frac(1-2sin(x)cos(x))(1+sin(x)cos(x))sin^2(x)-cos^2(x) \
& = frac1-sin(x)cos(x)-2sin^2(x)cos^2(x)sin^2(x)-cos^2(x) \
& = fracsin^2(x)sin^2(x)fraccsc^2(x)-cot(x)-2cos^2(x)1-cot^2(x).
endalign

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answered Jul 23 at 20:44

Math Lover

12.3k21232

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share|cite|improve this answer

answered Jul 23 at 20:44

Math Lover

12.3k21232

answered Jul 23 at 20:44

Math Lover

12.3k21232

answered Jul 23 at 20:44

Math Lover

12.3k21232

12.3k21232

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up vote
1
down vote

I'd start with the right-hand side:
beginalign
fraccsc^2x-cot x-2cos^2x1-cot^2x
&=fracdfrac1sin^2x-dfraccos xsin x-2cos^2x1-dfraccos^2xsin^2x
\[6px]
&=frac1-sin xcos x-2cos^2xsin^2xsin^2x-cos^2x
\[6px]
&=frac(sin^2x+cos^2x)^2-(sin^2x+cos^2x)sin xcos x-2sin^2xcos^2xsin^2x-cos^2x
\[6px]
&=fracsin^4x-sin^3xcos x-sin xcos^3x+cos^4xsin^2x-cos^2x
\[6px]
&=fracsin^3x(sin x-cos x)-cos^3x(sin x-cos x)sin^2x-cos^2x
\[6px]
&=frac(sin^3x-cos^3x)(sin x-cos x)(sin x+cos x)(sin x-cos x)
\[6px]
&=fracsin^3x-cos^3xsin x+cos x
endalign
The trick is to “make the numerator homogeneous”.

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answered Jul 23 at 21:01

egreg

164k1180187

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up vote
1
down vote

I'd start with the right-hand side:
beginalign
fraccsc^2x-cot x-2cos^2x1-cot^2x
&=fracdfrac1sin^2x-dfraccos xsin x-2cos^2x1-dfraccos^2xsin^2x
\[6px]
&=frac1-sin xcos x-2cos^2xsin^2xsin^2x-cos^2x
\[6px]
&=frac(sin^2x+cos^2x)^2-(sin^2x+cos^2x)sin xcos x-2sin^2xcos^2xsin^2x-cos^2x
\[6px]
&=fracsin^4x-sin^3xcos x-sin xcos^3x+cos^4xsin^2x-cos^2x
\[6px]
&=fracsin^3x(sin x-cos x)-cos^3x(sin x-cos x)sin^2x-cos^2x
\[6px]
&=frac(sin^3x-cos^3x)(sin x-cos x)(sin x+cos x)(sin x-cos x)
\[6px]
&=fracsin^3x-cos^3xsin x+cos x
endalign
The trick is to “make the numerator homogeneous”.

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answered Jul 23 at 21:01

egreg

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up vote
1
down vote

up vote
1
down vote

I'd start with the right-hand side:
beginalign
fraccsc^2x-cot x-2cos^2x1-cot^2x
&=fracdfrac1sin^2x-dfraccos xsin x-2cos^2x1-dfraccos^2xsin^2x
\[6px]
&=frac1-sin xcos x-2cos^2xsin^2xsin^2x-cos^2x
\[6px]
&=frac(sin^2x+cos^2x)^2-(sin^2x+cos^2x)sin xcos x-2sin^2xcos^2xsin^2x-cos^2x
\[6px]
&=fracsin^4x-sin^3xcos x-sin xcos^3x+cos^4xsin^2x-cos^2x
\[6px]
&=fracsin^3x(sin x-cos x)-cos^3x(sin x-cos x)sin^2x-cos^2x
\[6px]
&=frac(sin^3x-cos^3x)(sin x-cos x)(sin x+cos x)(sin x-cos x)
\[6px]
&=fracsin^3x-cos^3xsin x+cos x
endalign
The trick is to “make the numerator homogeneous”.

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answered Jul 23 at 21:01

egreg

164k1180187

I'd start with the right-hand side:
beginalign
fraccsc^2x-cot x-2cos^2x1-cot^2x
&=fracdfrac1sin^2x-dfraccos xsin x-2cos^2x1-dfraccos^2xsin^2x
\[6px]
&=frac1-sin xcos x-2cos^2xsin^2xsin^2x-cos^2x
\[6px]
&=frac(sin^2x+cos^2x)^2-(sin^2x+cos^2x)sin xcos x-2sin^2xcos^2xsin^2x-cos^2x
\[6px]
&=fracsin^4x-sin^3xcos x-sin xcos^3x+cos^4xsin^2x-cos^2x
\[6px]
&=fracsin^3x(sin x-cos x)-cos^3x(sin x-cos x)sin^2x-cos^2x
\[6px]
&=frac(sin^3x-cos^3x)(sin x-cos x)(sin x+cos x)(sin x-cos x)
\[6px]
&=fracsin^3x-cos^3xsin x+cos x
endalign
The trick is to “make the numerator homogeneous”.

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answered Jul 23 at 21:01

egreg

164k1180187

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share|cite|improve this answer

answered Jul 23 at 21:01

egreg

164k1180187

answered Jul 23 at 21:01

egreg

164k1180187

answered Jul 23 at 21:01

egreg

164k1180187

164k1180187

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Use that $$sin^2(x)(1-cot(x)^2)=sin(x)^2-cos(x)^2$$
then we get

$$fracsin(x)^3-cos(x)^3sin(x)+cos(x)=frac2cos(x)^4-2cos(x)^2-sin(x)cos(x)+1(sin(x)+cos(x))(sin(x)-cos(x))$$
and then use that
$$sin(x)^3-cos(x)^3=(sin(x)-cos(x))(sin(x)^2+sin(x)cos(x)+cos(x)^2)$$

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edited Jul 23 at 20:46

answered Jul 23 at 20:40

Dr. Sonnhard Graubner

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up vote
0
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Use that $$sin^2(x)(1-cot(x)^2)=sin(x)^2-cos(x)^2$$
then we get

$$fracsin(x)^3-cos(x)^3sin(x)+cos(x)=frac2cos(x)^4-2cos(x)^2-sin(x)cos(x)+1(sin(x)+cos(x))(sin(x)-cos(x))$$
and then use that
$$sin(x)^3-cos(x)^3=(sin(x)-cos(x))(sin(x)^2+sin(x)cos(x)+cos(x)^2)$$

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edited Jul 23 at 20:46

answered Jul 23 at 20:40

Dr. Sonnhard Graubner

66.7k32659

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up vote
0
down vote

up vote
0
down vote

Use that $$sin^2(x)(1-cot(x)^2)=sin(x)^2-cos(x)^2$$
then we get

$$fracsin(x)^3-cos(x)^3sin(x)+cos(x)=frac2cos(x)^4-2cos(x)^2-sin(x)cos(x)+1(sin(x)+cos(x))(sin(x)-cos(x))$$
and then use that
$$sin(x)^3-cos(x)^3=(sin(x)-cos(x))(sin(x)^2+sin(x)cos(x)+cos(x)^2)$$

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edited Jul 23 at 20:46

answered Jul 23 at 20:40

Dr. Sonnhard Graubner

66.7k32659

Use that $$sin^2(x)(1-cot(x)^2)=sin(x)^2-cos(x)^2$$
then we get

$$fracsin(x)^3-cos(x)^3sin(x)+cos(x)=frac2cos(x)^4-2cos(x)^2-sin(x)cos(x)+1(sin(x)+cos(x))(sin(x)-cos(x))$$
and then use that
$$sin(x)^3-cos(x)^3=(sin(x)-cos(x))(sin(x)^2+sin(x)cos(x)+cos(x)^2)$$

share|cite|improve this answer

edited Jul 23 at 20:46

answered Jul 23 at 20:40

Dr. Sonnhard Graubner

66.7k32659

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 20:46

edited Jul 23 at 20:46

edited Jul 23 at 20:46

answered Jul 23 at 20:40

Dr. Sonnhard Graubner

66.7k32659

answered Jul 23 at 20:40

Dr. Sonnhard Graubner

66.7k32659

answered Jul 23 at 20:40

Dr. Sonnhard Graubner

66.7k32659

66.7k32659

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