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Slither Like a Snake

Solving $ sin2xsin x+ cos^2x = sin5xsin 4x+ cos^24x$

2

$begingroup$

Solve the following equation over the real numbers:
$$ sin(2x)sin(x)+ cos^2(x) = sin(5x)sin(4x)+ cos^2(4x) $$

What I've tried so far, to no avail, is using identities (product-to-sum and sum-to-product, as well as the double angle identity) in search of a more convenient form, like a product equaling $0$ or a polynomial through some notation. Alternate forms I've reached :
$$ sin(5x)sin(3x)=sin(6x)sin (3x)$$
$$sin(x)(sin(x)(2cos(x)-1)- 4cos(x)(1-2sin^2(x))(sin(5x)-sin(4x) ))=0$$
Perhaps a helpful identity: $$ cos^2(x)- cos^2(4x)= sin(5x)sin(3x) $$

algebra-precalculus trigonometry

share|cite|improve this question

edited Apr 9 at 7:59

Blue

49.7k870158

asked Apr 8 at 14:14

Luca PanaLuca Pana

1297

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  use $cos^2(a)+sin^2(a) = 1$ first on both sides then factorize. Now use $sin(A)-sin(B)$ as a product on both sides and simplify (giving some roots), then write $sin(C)cos(D)$ as a difference of sines on both sides and simplify. finish with a difference of sines written as a product and solve.
  $endgroup$
  – Paul
  Apr 8 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've difficulties understanding your suggestion
  $endgroup$
  – Luca Pana
  Apr 8 at 14:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Give the first step a go though to replace the $cos^2$ terms then see where it goes..
  $endgroup$
  – Paul
  Apr 8 at 14:31
  

  
  
  
  

add a comment | 

2

$begingroup$

Solve the following equation over the real numbers:
$$ sin(2x)sin(x)+ cos^2(x) = sin(5x)sin(4x)+ cos^2(4x) $$

What I've tried so far, to no avail, is using identities (product-to-sum and sum-to-product, as well as the double angle identity) in search of a more convenient form, like a product equaling $0$ or a polynomial through some notation. Alternate forms I've reached :
$$ sin(5x)sin(3x)=sin(6x)sin (3x)$$
$$sin(x)(sin(x)(2cos(x)-1)- 4cos(x)(1-2sin^2(x))(sin(5x)-sin(4x) ))=0$$
Perhaps a helpful identity: $$ cos^2(x)- cos^2(4x)= sin(5x)sin(3x) $$

algebra-precalculus trigonometry

share|cite|improve this question

edited Apr 9 at 7:59

Blue

49.7k870158

asked Apr 8 at 14:14

Luca PanaLuca Pana

1297

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  use $cos^2(a)+sin^2(a) = 1$ first on both sides then factorize. Now use $sin(A)-sin(B)$ as a product on both sides and simplify (giving some roots), then write $sin(C)cos(D)$ as a difference of sines on both sides and simplify. finish with a difference of sines written as a product and solve.
  $endgroup$
  – Paul
  Apr 8 at 14:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've difficulties understanding your suggestion
  $endgroup$
  – Luca Pana
  Apr 8 at 14:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Give the first step a go though to replace the $cos^2$ terms then see where it goes..
  $endgroup$
  – Paul
  Apr 8 at 14:31
  

  
  
  
  

add a comment | 

$begingroup$

Solve the following equation over the real numbers:
$$ sin(2x)sin(x)+ cos^2(x) = sin(5x)sin(4x)+ cos^2(4x) $$

What I've tried so far, to no avail, is using identities (product-to-sum and sum-to-product, as well as the double angle identity) in search of a more convenient form, like a product equaling $0$ or a polynomial through some notation. Alternate forms I've reached :
$$ sin(5x)sin(3x)=sin(6x)sin (3x)$$
$$sin(x)(sin(x)(2cos(x)-1)- 4cos(x)(1-2sin^2(x))(sin(5x)-sin(4x) ))=0$$
Perhaps a helpful identity: $$ cos^2(x)- cos^2(4x)= sin(5x)sin(3x) $$

algebra-precalculus trigonometry

share|cite|improve this question

edited Apr 9 at 7:59

Blue

49.7k870158

asked Apr 8 at 14:14

Luca PanaLuca Pana

1297

$endgroup$

share|cite|improve this question

edited Apr 9 at 7:59

Blue

49.7k870158

asked Apr 8 at 14:14

Luca PanaLuca Pana

1297

share|cite|improve this question

edited Apr 9 at 7:59

Blue

49.7k870158

asked Apr 8 at 14:14

Luca PanaLuca Pana

1297

edited Apr 9 at 7:59

Blue

49.7k870158

edited Apr 9 at 7:59

Blue

49.7k870158

asked Apr 8 at 14:14

Luca PanaLuca Pana

1297

asked Apr 8 at 14:14

Luca PanaLuca Pana

1297

2 Answers
2

active

oldest

votes

0

$begingroup$

Hint

By $cos2y=2cos^2y-1$ and

http://mathworld.wolfram.com/WernerFormulas.html and

http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html,

$$cos x-cos3x+1+cos2x=cos x-cos9x+1+cos8x$$

$$cos2x-cos8x=cos3x-cos9x$$

$$2sin5xsin3x=2sin3xsin6x$$

share|cite|improve this answer

answered Apr 8 at 14:31

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes this was an alternate form I reached, when posting I messed up a sign
  $endgroup$
  – Luca Pana
  Apr 8 at 14:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Luca, So we are done?
  $endgroup$
  – lab bhattacharjee
  Apr 8 at 14:39
  

  
  
  
  

add a comment | 

0

$begingroup$

use $cos^2(a)+sin^2(a)=1$ first on both sides then factorize.

$$sin(x)[sin(2x)-sin(x)]=sin(4x)[sin(5x)-sin(4x)]$$

Now use sin(A)−sin(B) as a product on both sides and simplify (giving some roots $sin(fracx2)=0$)

$$sin(x).2cos(frac3x2)sin(fracx2)=sin(4x).2cos(frac9x2)sin(fracx2)$$

then write sin(C)cos(D) as a difference of sines on both sides and simplify.

$$frac12[sin(frac5x2)+sin(frac-x2)]=frac12[sin(frac17x2)+sin(frac-x2)]$$

Write the difference of sines as a product and solve

$$sin(frac5x2)-sin(frac17x2)=0$$

share|cite|improve this answer

answered Apr 9 at 7:16

PaulPaul

97076

$endgroup$

add a comment | 

Your Answer

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0

$begingroup$

Hint

By $cos2y=2cos^2y-1$ and

http://mathworld.wolfram.com/WernerFormulas.html and

http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html,

$$cos x-cos3x+1+cos2x=cos x-cos9x+1+cos8x$$

$$cos2x-cos8x=cos3x-cos9x$$

$$2sin5xsin3x=2sin3xsin6x$$

share|cite|improve this answer

answered Apr 8 at 14:31

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes this was an alternate form I reached, when posting I messed up a sign
  $endgroup$
  – Luca Pana
  Apr 8 at 14:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Luca, So we are done?
  $endgroup$
  – lab bhattacharjee
  Apr 8 at 14:39
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint

By $cos2y=2cos^2y-1$ and

http://mathworld.wolfram.com/WernerFormulas.html and

http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html,

$$cos x-cos3x+1+cos2x=cos x-cos9x+1+cos8x$$

$$cos2x-cos8x=cos3x-cos9x$$

$$2sin5xsin3x=2sin3xsin6x$$

share|cite|improve this answer

answered Apr 8 at 14:31

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes this was an alternate form I reached, when posting I messed up a sign
  $endgroup$
  – Luca Pana
  Apr 8 at 14:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Luca, So we are done?
  $endgroup$
  – lab bhattacharjee
  Apr 8 at 14:39
  

  
  
  
  

add a comment | 

$begingroup$

Hint

By $cos2y=2cos^2y-1$ and

http://mathworld.wolfram.com/WernerFormulas.html and

http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html,

$$cos x-cos3x+1+cos2x=cos x-cos9x+1+cos8x$$

$$cos2x-cos8x=cos3x-cos9x$$

$$2sin5xsin3x=2sin3xsin6x$$

share|cite|improve this answer

answered Apr 8 at 14:31

lab bhattacharjeelab bhattacharjee

228k15159279

$endgroup$

share|cite|improve this answer

answered Apr 8 at 14:31

lab bhattacharjeelab bhattacharjee

228k15159279

answered Apr 8 at 14:31

lab bhattacharjeelab bhattacharjee

228k15159279

answered Apr 8 at 14:31

lab bhattacharjeelab bhattacharjee

228k15159279

0

$begingroup$

use $cos^2(a)+sin^2(a)=1$ first on both sides then factorize.

$$sin(x)[sin(2x)-sin(x)]=sin(4x)[sin(5x)-sin(4x)]$$

Now use sin(A)−sin(B) as a product on both sides and simplify (giving some roots $sin(fracx2)=0$)

$$sin(x).2cos(frac3x2)sin(fracx2)=sin(4x).2cos(frac9x2)sin(fracx2)$$

then write sin(C)cos(D) as a difference of sines on both sides and simplify.

$$frac12[sin(frac5x2)+sin(frac-x2)]=frac12[sin(frac17x2)+sin(frac-x2)]$$

Write the difference of sines as a product and solve

$$sin(frac5x2)-sin(frac17x2)=0$$

share|cite|improve this answer

answered Apr 9 at 7:16

PaulPaul

97076

$endgroup$

add a comment | 

0

$begingroup$

use $cos^2(a)+sin^2(a)=1$ first on both sides then factorize.

$$sin(x)[sin(2x)-sin(x)]=sin(4x)[sin(5x)-sin(4x)]$$

Now use sin(A)−sin(B) as a product on both sides and simplify (giving some roots $sin(fracx2)=0$)

$$sin(x).2cos(frac3x2)sin(fracx2)=sin(4x).2cos(frac9x2)sin(fracx2)$$

then write sin(C)cos(D) as a difference of sines on both sides and simplify.

$$frac12[sin(frac5x2)+sin(frac-x2)]=frac12[sin(frac17x2)+sin(frac-x2)]$$

Write the difference of sines as a product and solve

$$sin(frac5x2)-sin(frac17x2)=0$$

share|cite|improve this answer

answered Apr 9 at 7:16

PaulPaul

97076

$endgroup$

add a comment | 

$begingroup$

use $cos^2(a)+sin^2(a)=1$ first on both sides then factorize.

$$sin(x)[sin(2x)-sin(x)]=sin(4x)[sin(5x)-sin(4x)]$$

Now use sin(A)−sin(B) as a product on both sides and simplify (giving some roots $sin(fracx2)=0$)

$$sin(x).2cos(frac3x2)sin(fracx2)=sin(4x).2cos(frac9x2)sin(fracx2)$$

then write sin(C)cos(D) as a difference of sines on both sides and simplify.

$$frac12[sin(frac5x2)+sin(frac-x2)]=frac12[sin(frac17x2)+sin(frac-x2)]$$

Write the difference of sines as a product and solve

$$sin(frac5x2)-sin(frac17x2)=0$$

share|cite|improve this answer

answered Apr 9 at 7:16

PaulPaul

97076

$endgroup$

share|cite|improve this answer

answered Apr 9 at 7:16

PaulPaul

97076

answered Apr 9 at 7:16

PaulPaul

97076

answered Apr 9 at 7:16

PaulPaul

97076