$[-sin(s)+4tsin(2s)][-2sin(2s)] + [2cos(2s)][cos(s)-4tcos(2s)] = 0$ $rightarrow t=frac{1}{4}cos(s)$ [closed]
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$[-sin(s)+4tsin(2s)][-2sin(2s)] + [2cos(2s)][cos(s)-4tcos(2s)] = 0$ $rightarrow t=frac{1}{4}cos(s)$
I cant seem to prove this
trigonometry
edited Dec 10 '18 at 15:09
Andrei
12.1k21126
asked Dec 10 '18 at 15:06
pablo_mathscobarpablo_mathscobar
996
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closed as off-topic by Saad, Arnaud D., Harry49, Robert Z, TMM Dec 12 '18 at 16:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
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$begingroup$
$[-sin(s)+4tsin(2s)][-2sin(2s)] + [2cos(2s)][cos(s)-4tcos(2s)] = 0$ $rightarrow t=frac{1}{4}cos(s)$
I cant seem to prove this
trigonometry
edited Dec 10 '18 at 15:09
Andrei
12.1k21126
asked Dec 10 '18 at 15:06
pablo_mathscobarpablo_mathscobar
996
$endgroup$
closed as off-topic by Saad, Arnaud D., Harry49, Robert Z, TMM Dec 12 '18 at 16:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Arnaud D., Harry49, Robert Z, TMM
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$[-sin(s)+4tsin(2s)][-2sin(2s)] + [2cos(2s)][cos(s)-4tcos(2s)] = 0$ $rightarrow t=frac{1}{4}cos(s)$
I cant seem to prove this
trigonometry
edited Dec 10 '18 at 15:09
Andrei
12.1k21126
asked Dec 10 '18 at 15:06
pablo_mathscobarpablo_mathscobar
996
$endgroup$
$[-sin(s)+4tsin(2s)][-2sin(2s)] + [2cos(2s)][cos(s)-4tcos(2s)] = 0$ $rightarrow t=frac{1}{4}cos(s)$
I cant seem to prove this
trigonometry
trigonometry
edited Dec 10 '18 at 15:09
Andrei
12.1k21126
asked Dec 10 '18 at 15:06
pablo_mathscobarpablo_mathscobar
996
edited Dec 10 '18 at 15:09
Andrei
12.1k21126
asked Dec 10 '18 at 15:06
pablo_mathscobarpablo_mathscobar
996
edited Dec 10 '18 at 15:09
Andrei
12.1k21126
edited Dec 10 '18 at 15:09
Andrei
12.1k21126
edited Dec 10 '18 at 15:09
Andrei
12.1k21126
12.1k21126
asked Dec 10 '18 at 15:06
pablo_mathscobarpablo_mathscobar
996
asked Dec 10 '18 at 15:06
pablo_mathscobarpablo_mathscobar
996
asked Dec 10 '18 at 15:06
pablo_mathscobarpablo_mathscobar
996
996
closed as off-topic by Saad, Arnaud D., Harry49, Robert Z, TMM Dec 12 '18 at 16:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Arnaud D., Harry49, Robert Z, TMM
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Arnaud D., Harry49, Robert Z, TMM Dec 12 '18 at 16:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Arnaud D., Harry49, Robert Z, TMM
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
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Group together the terms with t. You will get something proportional to $cos^2 2s+sin^2 2s$. The rest of the terms are of type $cos a cos b+sin a sin b$
answered Dec 10 '18 at 15:14
AndreiAndrei
12.1k21126
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Hint:
Elementary, Watson!
$cos(2s-s)=?$
$$cos^22s+sin^22s=?$$
answered Dec 10 '18 at 15:12
lab bhattacharjeelab bhattacharjee
225k15157275
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Opening the brackets and multiplying we get
2sinS sin2S-8t sin^2 2S + 2cos 2S cos S- 8tcos^2 2S =0
Implies
2(sin2S sin S+ cos2S cosS) - 8t(sin^2 2S+cos^2 2S) =0
ie, 2cos(2S-S)-8t*1 =0
Which gives 2cosS - 8t=0
2cosS =8t
t= 1/4 cost
answered Dec 10 '18 at 15:42
ChinnuChinnu
11
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Group together the terms with t. You will get something proportional to $cos^2 2s+sin^2 2s$. The rest of the terms are of type $cos a cos b+sin a sin b$
answered Dec 10 '18 at 15:14
AndreiAndrei
12.1k21126
$endgroup$
add a comment |
$begingroup$
Group together the terms with t. You will get something proportional to $cos^2 2s+sin^2 2s$. The rest of the terms are of type $cos a cos b+sin a sin b$
answered Dec 10 '18 at 15:14
AndreiAndrei
12.1k21126
$endgroup$
add a comment |
$begingroup$
Group together the terms with t. You will get something proportional to $cos^2 2s+sin^2 2s$. The rest of the terms are of type $cos a cos b+sin a sin b$
answered Dec 10 '18 at 15:14
AndreiAndrei
12.1k21126
$endgroup$
Group together the terms with t. You will get something proportional to $cos^2 2s+sin^2 2s$. The rest of the terms are of type $cos a cos b+sin a sin b$
answered Dec 10 '18 at 15:14
AndreiAndrei
12.1k21126
answered Dec 10 '18 at 15:14
AndreiAndrei
12.1k21126
answered Dec 10 '18 at 15:14
AndreiAndrei
12.1k21126
answered Dec 10 '18 at 15:14
AndreiAndrei
12.1k21126
12.1k21126
add a comment |
add a comment |
$begingroup$
Hint:
Elementary, Watson!
$cos(2s-s)=?$
$$cos^22s+sin^22s=?$$
answered Dec 10 '18 at 15:12
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
Hint:
Elementary, Watson!
$cos(2s-s)=?$
$$cos^22s+sin^22s=?$$
answered Dec 10 '18 at 15:12
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
add a comment |
$begingroup$
Hint:
Elementary, Watson!
$cos(2s-s)=?$
$$cos^22s+sin^22s=?$$
answered Dec 10 '18 at 15:12
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
Hint:
Elementary, Watson!
$cos(2s-s)=?$
$$cos^22s+sin^22s=?$$
answered Dec 10 '18 at 15:12
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 10 '18 at 15:12
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 10 '18 at 15:12
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 10 '18 at 15:12
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
add a comment |
add a comment |
$begingroup$
Opening the brackets and multiplying we get
2sinS sin2S-8t sin^2 2S + 2cos 2S cos S- 8tcos^2 2S =0
Implies
2(sin2S sin S+ cos2S cosS) - 8t(sin^2 2S+cos^2 2S) =0
ie, 2cos(2S-S)-8t*1 =0
Which gives 2cosS - 8t=0
2cosS =8t
t= 1/4 cost
answered Dec 10 '18 at 15:42
ChinnuChinnu
11
$endgroup$
add a comment |
$begingroup$
Opening the brackets and multiplying we get
2sinS sin2S-8t sin^2 2S + 2cos 2S cos S- 8tcos^2 2S =0
Implies
2(sin2S sin S+ cos2S cosS) - 8t(sin^2 2S+cos^2 2S) =0
ie, 2cos(2S-S)-8t*1 =0
Which gives 2cosS - 8t=0
2cosS =8t
t= 1/4 cost
answered Dec 10 '18 at 15:42
ChinnuChinnu
11
$endgroup$
add a comment |
$begingroup$
Opening the brackets and multiplying we get
2sinS sin2S-8t sin^2 2S + 2cos 2S cos S- 8tcos^2 2S =0
Implies
2(sin2S sin S+ cos2S cosS) - 8t(sin^2 2S+cos^2 2S) =0
ie, 2cos(2S-S)-8t*1 =0
Which gives 2cosS - 8t=0
2cosS =8t
t= 1/4 cost
answered Dec 10 '18 at 15:42
ChinnuChinnu
11
$endgroup$
Opening the brackets and multiplying we get
2sinS sin2S-8t sin^2 2S + 2cos 2S cos S- 8tcos^2 2S =0
Implies
2(sin2S sin S+ cos2S cosS) - 8t(sin^2 2S+cos^2 2S) =0
ie, 2cos(2S-S)-8t*1 =0
Which gives 2cosS - 8t=0
2cosS =8t
t= 1/4 cost
answered Dec 10 '18 at 15:42
ChinnuChinnu
11
answered Dec 10 '18 at 15:42
ChinnuChinnu
11
answered Dec 10 '18 at 15:42
ChinnuChinnu
11
answered Dec 10 '18 at 15:42
ChinnuChinnu
11
11
add a comment |
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