Evaluating $dfrac1sin(2x) + dfrac1sin(4x) + dfrac1sin(8x) + dfrac1sin(16x)$
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Evaluate
$$dfrac1sin(2x) + dfrac1sin(4x) + dfrac1sin(8x) + dfrac1sin(16x)$$
It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.
Rewriting and using trigonometric identities
$$dfrac1sin(2x) + dfrac1sin(2x) cos (2x) + dfrac1 2big [2sin (2x)cos (2x)cos (4x)big ] + dfrac1sin(16x)$$
What am I missing?
Regards
trigonometry
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up vote
2
down vote
favorite
Evaluate
$$dfrac1sin(2x) + dfrac1sin(4x) + dfrac1sin(8x) + dfrac1sin(16x)$$
It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.
Rewriting and using trigonometric identities
$$dfrac1sin(2x) + dfrac1sin(2x) cos (2x) + dfrac1 2big [2sin (2x)cos (2x)cos (4x)big ] + dfrac1sin(16x)$$
What am I missing?
Regards
trigonometry
See math.stackexchange.com/questions/1591220/â¦
â lab bhattacharjee
Sep 6 at 11:27
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
Evaluate
$$dfrac1sin(2x) + dfrac1sin(4x) + dfrac1sin(8x) + dfrac1sin(16x)$$
It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.
Rewriting and using trigonometric identities
$$dfrac1sin(2x) + dfrac1sin(2x) cos (2x) + dfrac1 2big [2sin (2x)cos (2x)cos (4x)big ] + dfrac1sin(16x)$$
What am I missing?
Regards
trigonometry
Evaluate
$$dfrac1sin(2x) + dfrac1sin(4x) + dfrac1sin(8x) + dfrac1sin(16x)$$
It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.
Rewriting and using trigonometric identities
$$dfrac1sin(2x) + dfrac1sin(2x) cos (2x) + dfrac1 2big [2sin (2x)cos (2x)cos (4x)big ] + dfrac1sin(16x)$$
What am I missing?
Regards
trigonometry
trigonometry
asked Sep 6 at 11:11
Busi
325110
325110
See math.stackexchange.com/questions/1591220/â¦
â lab bhattacharjee
Sep 6 at 11:27
add a comment |Â
See math.stackexchange.com/questions/1591220/â¦
â lab bhattacharjee
Sep 6 at 11:27
See math.stackexchange.com/questions/1591220/â¦
â lab bhattacharjee
Sep 6 at 11:27
See math.stackexchange.com/questions/1591220/â¦
â lab bhattacharjee
Sep 6 at 11:27
add a comment |Â
3 Answers
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$$sin(A-B)=sin A cos B-cos A sin B$$
$$frac1sin 2x=dfracsin(2x-x)sin2xsin x=cot x-cot2x$$
$$frac1sin 4x=dfracsin(4x-2x)sin4xsin 2x=cot 2x-cot4x$$
$$frac1sin 8x=dfracsin(8x-4x)sin8xsin 4x=cot 4x-cot8x$$
$$frac1sin 16x=dfracsin(16x-8x)sin16xsin 8x=cot 8x-cot16x$$
Where did that $cot$ come from?
â Busi
Sep 6 at 12:23
apply $sin(a-b)$ formula
â Deepesh Meena
Sep 6 at 13:42
@DeepeshMeena, Have you noticed the link?
â lab bhattacharjee
Sep 7 at 7:44
which link can you provide me?
â Deepesh Meena
Sep 7 at 9:42
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up vote
2
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$$frac1sin2x=fracsin xsin 2x sin x=fracsin (2x-x)sin 2x sin x=fracsin 2x cos x - cos 2x sin xsin 2x sin x=cot x - cot2x$$
$$frac1sin4x=cot 2x - cot4x$$
$$frac1sin8x=cot 4x - cot8x$$
$$frac1sin16x=cot 8x - cot16x$$
$$frac1sin2x+frac1sin4x+frac1sin8x+frac1sin16x=cot x-cot 16x$$
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Hint : Notice that $frac1sin(2x))$ is a common factor of all terms of the sum.
Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
â PackSciences
Sep 6 at 11:26
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$$sin(A-B)=sin A cos B-cos A sin B$$
$$frac1sin 2x=dfracsin(2x-x)sin2xsin x=cot x-cot2x$$
$$frac1sin 4x=dfracsin(4x-2x)sin4xsin 2x=cot 2x-cot4x$$
$$frac1sin 8x=dfracsin(8x-4x)sin8xsin 4x=cot 4x-cot8x$$
$$frac1sin 16x=dfracsin(16x-8x)sin16xsin 8x=cot 8x-cot16x$$
Where did that $cot$ come from?
â Busi
Sep 6 at 12:23
apply $sin(a-b)$ formula
â Deepesh Meena
Sep 6 at 13:42
@DeepeshMeena, Have you noticed the link?
â lab bhattacharjee
Sep 7 at 7:44
which link can you provide me?
â Deepesh Meena
Sep 7 at 9:42
add a comment |Â
up vote
3
down vote
$$sin(A-B)=sin A cos B-cos A sin B$$
$$frac1sin 2x=dfracsin(2x-x)sin2xsin x=cot x-cot2x$$
$$frac1sin 4x=dfracsin(4x-2x)sin4xsin 2x=cot 2x-cot4x$$
$$frac1sin 8x=dfracsin(8x-4x)sin8xsin 4x=cot 4x-cot8x$$
$$frac1sin 16x=dfracsin(16x-8x)sin16xsin 8x=cot 8x-cot16x$$
Where did that $cot$ come from?
â Busi
Sep 6 at 12:23
apply $sin(a-b)$ formula
â Deepesh Meena
Sep 6 at 13:42
@DeepeshMeena, Have you noticed the link?
â lab bhattacharjee
Sep 7 at 7:44
which link can you provide me?
â Deepesh Meena
Sep 7 at 9:42
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$$sin(A-B)=sin A cos B-cos A sin B$$
$$frac1sin 2x=dfracsin(2x-x)sin2xsin x=cot x-cot2x$$
$$frac1sin 4x=dfracsin(4x-2x)sin4xsin 2x=cot 2x-cot4x$$
$$frac1sin 8x=dfracsin(8x-4x)sin8xsin 4x=cot 4x-cot8x$$
$$frac1sin 16x=dfracsin(16x-8x)sin16xsin 8x=cot 8x-cot16x$$
$$sin(A-B)=sin A cos B-cos A sin B$$
$$frac1sin 2x=dfracsin(2x-x)sin2xsin x=cot x-cot2x$$
$$frac1sin 4x=dfracsin(4x-2x)sin4xsin 2x=cot 2x-cot4x$$
$$frac1sin 8x=dfracsin(8x-4x)sin8xsin 4x=cot 4x-cot8x$$
$$frac1sin 16x=dfracsin(16x-8x)sin16xsin 8x=cot 8x-cot16x$$
edited Sep 15 at 0:13
answered Sep 6 at 12:04
Deepesh Meena
3,8912825
3,8912825
Where did that $cot$ come from?
â Busi
Sep 6 at 12:23
apply $sin(a-b)$ formula
â Deepesh Meena
Sep 6 at 13:42
@DeepeshMeena, Have you noticed the link?
â lab bhattacharjee
Sep 7 at 7:44
which link can you provide me?
â Deepesh Meena
Sep 7 at 9:42
add a comment |Â
Where did that $cot$ come from?
â Busi
Sep 6 at 12:23
apply $sin(a-b)$ formula
â Deepesh Meena
Sep 6 at 13:42
@DeepeshMeena, Have you noticed the link?
â lab bhattacharjee
Sep 7 at 7:44
which link can you provide me?
â Deepesh Meena
Sep 7 at 9:42
Where did that $cot$ come from?
â Busi
Sep 6 at 12:23
Where did that $cot$ come from?
â Busi
Sep 6 at 12:23
apply $sin(a-b)$ formula
â Deepesh Meena
Sep 6 at 13:42
apply $sin(a-b)$ formula
â Deepesh Meena
Sep 6 at 13:42
@DeepeshMeena, Have you noticed the link?
â lab bhattacharjee
Sep 7 at 7:44
@DeepeshMeena, Have you noticed the link?
â lab bhattacharjee
Sep 7 at 7:44
which link can you provide me?
â Deepesh Meena
Sep 7 at 9:42
which link can you provide me?
â Deepesh Meena
Sep 7 at 9:42
add a comment |Â
up vote
2
down vote
$$frac1sin2x=fracsin xsin 2x sin x=fracsin (2x-x)sin 2x sin x=fracsin 2x cos x - cos 2x sin xsin 2x sin x=cot x - cot2x$$
$$frac1sin4x=cot 2x - cot4x$$
$$frac1sin8x=cot 4x - cot8x$$
$$frac1sin16x=cot 8x - cot16x$$
$$frac1sin2x+frac1sin4x+frac1sin8x+frac1sin16x=cot x-cot 16x$$
add a comment |Â
up vote
2
down vote
$$frac1sin2x=fracsin xsin 2x sin x=fracsin (2x-x)sin 2x sin x=fracsin 2x cos x - cos 2x sin xsin 2x sin x=cot x - cot2x$$
$$frac1sin4x=cot 2x - cot4x$$
$$frac1sin8x=cot 4x - cot8x$$
$$frac1sin16x=cot 8x - cot16x$$
$$frac1sin2x+frac1sin4x+frac1sin8x+frac1sin16x=cot x-cot 16x$$
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$frac1sin2x=fracsin xsin 2x sin x=fracsin (2x-x)sin 2x sin x=fracsin 2x cos x - cos 2x sin xsin 2x sin x=cot x - cot2x$$
$$frac1sin4x=cot 2x - cot4x$$
$$frac1sin8x=cot 4x - cot8x$$
$$frac1sin16x=cot 8x - cot16x$$
$$frac1sin2x+frac1sin4x+frac1sin8x+frac1sin16x=cot x-cot 16x$$
$$frac1sin2x=fracsin xsin 2x sin x=fracsin (2x-x)sin 2x sin x=fracsin 2x cos x - cos 2x sin xsin 2x sin x=cot x - cot2x$$
$$frac1sin4x=cot 2x - cot4x$$
$$frac1sin8x=cot 4x - cot8x$$
$$frac1sin16x=cot 8x - cot16x$$
$$frac1sin2x+frac1sin4x+frac1sin8x+frac1sin16x=cot x-cot 16x$$
answered Sep 6 at 12:08
Oldboy
3,3201323
3,3201323
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint : Notice that $frac1sin(2x))$ is a common factor of all terms of the sum.
Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
â PackSciences
Sep 6 at 11:26
add a comment |Â
up vote
1
down vote
Hint : Notice that $frac1sin(2x))$ is a common factor of all terms of the sum.
Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
â PackSciences
Sep 6 at 11:26
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint : Notice that $frac1sin(2x))$ is a common factor of all terms of the sum.
Hint : Notice that $frac1sin(2x))$ is a common factor of all terms of the sum.
answered Sep 6 at 11:19
PackSciences
41616
41616
Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
â PackSciences
Sep 6 at 11:26
add a comment |Â
Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
â PackSciences
Sep 6 at 11:26
Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
â PackSciences
Sep 6 at 11:26
Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
â PackSciences
Sep 6 at 11:26
add a comment |Â
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See math.stackexchange.com/questions/1591220/â¦
â lab bhattacharjee
Sep 6 at 11:27