Show that $2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=0$ as...
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Show that $2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=0$ as simply as possible. [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)How to get $ cot(theta/2)$ from $ frac {sin theta} {1 - cos theta} $?If $sintheta+sinphi=a$ and $costheta+ cosphi=b$, then find $tan frac{theta-phi}2$.How does $sin theta times sin (theta - phi)$ become $frac{cos phi - cos (2theta-phi)}{2}$Verifying $frac{cos{2theta}}{1 + sin{2theta}} = frac{cot{theta} - 1}{cot{theta} + 1}$Prove $sin^2 theta +cos^4 theta =cos^2 theta +sin^4 theta $How would I establish identity $,dfrac{cottheta + sectheta}{costheta + tantheta} = sectheta cottheta,$?If $sintheta + sinphi = a$ and $costheta + cosphi = b$, then $sin(theta+phi) = ???$Prove that $frac{sin(x+theta)}{sin(x+phi)} = cos(theta - phi) +cot(x+phi)sin(theta-phi)$.Prove $sin(theta )+sin(theta -phi )=2cos(frac{phi }{2})sin(theta -frac{phi }{2})$Show $frac{cos(ntheta)-cos(n+1)theta)}{2-2costheta}=frac{sin(n+1/2)theta}{2sin(theta/2)}$
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Using the fewest possible number of trig identities, how do you show that
$quad2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=0$?
trigonometry vector-analysis
asked Mar 25 at 16:45
LewisLewis
33
$endgroup$
closed as off-topic by mrtaurho, John Omielan, Shailesh, zz20s, Parcly Taxel Mar 26 at 3:25
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." – mrtaurho, John Omielan, Shailesh, zz20s, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Using the fewest possible number of trig identities, how do you show that
$quad2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=0$?
trigonometry vector-analysis
asked Mar 25 at 16:45
LewisLewis
33
$endgroup$
closed as off-topic by mrtaurho, John Omielan, Shailesh, zz20s, Parcly Taxel Mar 26 at 3:25
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." – mrtaurho, John Omielan, Shailesh, zz20s, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I'd start with the non-trigonometric simplification of dividing-through by the common factor of $2sinphi$. (Assuming $sin phineq 0$, of course. But, then, if $sinphi=0$, the equality is trivial.) This reduces a lot of symbolic clutter, clearing the way for some actual trig.
$endgroup$
– Blue
Mar 25 at 16:54
add a comment |
$begingroup$
Using the fewest possible number of trig identities, how do you show that
$quad2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=0$?
trigonometry vector-analysis
asked Mar 25 at 16:45
LewisLewis
33
$endgroup$
Using the fewest possible number of trig identities, how do you show that
$quad2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=0$?
trigonometry vector-analysis
trigonometry vector-analysis
asked Mar 25 at 16:45
LewisLewis
33
asked Mar 25 at 16:45
LewisLewis
33
asked Mar 25 at 16:45
LewisLewis
33
asked Mar 25 at 16:45
LewisLewis
33
asked Mar 25 at 16:45
LewisLewis
33
33
closed as off-topic by mrtaurho, John Omielan, Shailesh, zz20s, Parcly Taxel Mar 26 at 3:25
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." – mrtaurho, John Omielan, Shailesh, zz20s, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by mrtaurho, John Omielan, Shailesh, zz20s, Parcly Taxel Mar 26 at 3:25
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." – mrtaurho, John Omielan, Shailesh, zz20s, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I'd start with the non-trigonometric simplification of dividing-through by the common factor of $2sinphi$. (Assuming $sin phineq 0$, of course. But, then, if $sinphi=0$, the equality is trivial.) This reduces a lot of symbolic clutter, clearing the way for some actual trig.
$endgroup$
– Blue
Mar 25 at 16:54
add a comment |
1
$begingroup$
I'd start with the non-trigonometric simplification of dividing-through by the common factor of $2sinphi$. (Assuming $sin phineq 0$, of course. But, then, if $sinphi=0$, the equality is trivial.) This reduces a lot of symbolic clutter, clearing the way for some actual trig.
$endgroup$
– Blue
Mar 25 at 16:54
1
1
$begingroup$
I'd start with the non-trigonometric simplification of dividing-through by the common factor of $2sinphi$. (Assuming $sin phineq 0$, of course. But, then, if $sinphi=0$, the equality is trivial.) This reduces a lot of symbolic clutter, clearing the way for some actual trig.
$endgroup$
– Blue
Mar 25 at 16:54
$begingroup$
I'd start with the non-trigonometric simplification of dividing-through by the common factor of $2sinphi$. (Assuming $sin phineq 0$, of course. But, then, if $sinphi=0$, the equality is trivial.) This reduces a lot of symbolic clutter, clearing the way for some actual trig.
$endgroup$
– Blue
Mar 25 at 16:54
add a comment |
5 Answers
5
active
oldest
votes
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Hint: Use $$cos(2theta) = cos^2theta - sin^2 theta$$
answered Mar 25 at 16:51
Haris GusicHaris Gusic
3,595627
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add a comment |
$begingroup$
You see that the only trigonometric function there that doesn't have a single angle is the middle term. In fact, you have a double-angle. Consider expanding that guy with the double angle identity for cosine and see your result magically vanish.
answered Mar 25 at 16:51
Andrés Franco ValienteAndrés Franco Valiente
111
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$begingroup$
Thanks. Sometimes it's hard to see what's right in front of you!
$endgroup$
– Lewis
Mar 26 at 10:15
add a comment |
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$$LHS$$ $=quad2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=frac{2sin{phi}sin^2{theta}-2sin{phi}sin^2{theta}+2sin{phi}cos^2{theta}-2sin{phi}cos^2{theta}}{sin{theta}}=frac{0}{sin{theta}}=0=RHS$
answered Mar 25 at 17:03
saket kumarsaket kumar
167113
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add a comment |
$begingroup$
Hint: It is $$2sin(phi)sin(theta)+frac{2sin(phi)cos(2theta)}{sin(theta)}=frac{2sin(phi)cos(theta)^2}{sin(theta)}$$
answered Mar 25 at 16:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.2k42867
$endgroup$
add a comment |
$begingroup$
Multiply by $dfrac{sintheta}{2sinphi}$, and get
$$sin^2{theta}+cos{(2theta)}-cos^2{theta}=0.$$
answered Mar 25 at 16:55
Yves DaoustYves Daoust
133k676232
$endgroup$
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Use $$cos(2theta) = cos^2theta - sin^2 theta$$
answered Mar 25 at 16:51
Haris GusicHaris Gusic
3,595627
$endgroup$
add a comment |
$begingroup$
Hint: Use $$cos(2theta) = cos^2theta - sin^2 theta$$
answered Mar 25 at 16:51
Haris GusicHaris Gusic
3,595627
$endgroup$
add a comment |
$begingroup$
Hint: Use $$cos(2theta) = cos^2theta - sin^2 theta$$
answered Mar 25 at 16:51
Haris GusicHaris Gusic
3,595627
$endgroup$
Hint: Use $$cos(2theta) = cos^2theta - sin^2 theta$$
answered Mar 25 at 16:51
Haris GusicHaris Gusic
3,595627
answered Mar 25 at 16:51
Haris GusicHaris Gusic
3,595627
answered Mar 25 at 16:51
Haris GusicHaris Gusic
3,595627
answered Mar 25 at 16:51
Haris GusicHaris Gusic
3,595627
3,595627
add a comment |
add a comment |
$begingroup$
You see that the only trigonometric function there that doesn't have a single angle is the middle term. In fact, you have a double-angle. Consider expanding that guy with the double angle identity for cosine and see your result magically vanish.
answered Mar 25 at 16:51
Andrés Franco ValienteAndrés Franco Valiente
111
$endgroup$
$begingroup$
Thanks. Sometimes it's hard to see what's right in front of you!
$endgroup$
– Lewis
Mar 26 at 10:15
add a comment |
$begingroup$
You see that the only trigonometric function there that doesn't have a single angle is the middle term. In fact, you have a double-angle. Consider expanding that guy with the double angle identity for cosine and see your result magically vanish.
answered Mar 25 at 16:51
Andrés Franco ValienteAndrés Franco Valiente
111
$endgroup$
$begingroup$
Thanks. Sometimes it's hard to see what's right in front of you!
$endgroup$
– Lewis
Mar 26 at 10:15
add a comment |
$begingroup$
You see that the only trigonometric function there that doesn't have a single angle is the middle term. In fact, you have a double-angle. Consider expanding that guy with the double angle identity for cosine and see your result magically vanish.
answered Mar 25 at 16:51
Andrés Franco ValienteAndrés Franco Valiente
111
$endgroup$
You see that the only trigonometric function there that doesn't have a single angle is the middle term. In fact, you have a double-angle. Consider expanding that guy with the double angle identity for cosine and see your result magically vanish.
answered Mar 25 at 16:51
Andrés Franco ValienteAndrés Franco Valiente
111
answered Mar 25 at 16:51
Andrés Franco ValienteAndrés Franco Valiente
111
answered Mar 25 at 16:51
Andrés Franco ValienteAndrés Franco Valiente
111
answered Mar 25 at 16:51
Andrés Franco ValienteAndrés Franco Valiente
111
111
$begingroup$
Thanks. Sometimes it's hard to see what's right in front of you!
$endgroup$
– Lewis
Mar 26 at 10:15
add a comment |
$begingroup$
Thanks. Sometimes it's hard to see what's right in front of you!
$endgroup$
– Lewis
Mar 26 at 10:15
$begingroup$
Thanks. Sometimes it's hard to see what's right in front of you!
$endgroup$
– Lewis
Mar 26 at 10:15
$begingroup$
Thanks. Sometimes it's hard to see what's right in front of you!
$endgroup$
– Lewis
Mar 26 at 10:15
add a comment |
$begingroup$
$$LHS$$ $=quad2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=frac{2sin{phi}sin^2{theta}-2sin{phi}sin^2{theta}+2sin{phi}cos^2{theta}-2sin{phi}cos^2{theta}}{sin{theta}}=frac{0}{sin{theta}}=0=RHS$
answered Mar 25 at 17:03
saket kumarsaket kumar
167113
$endgroup$
add a comment |
$begingroup$
$$LHS$$ $=quad2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=frac{2sin{phi}sin^2{theta}-2sin{phi}sin^2{theta}+2sin{phi}cos^2{theta}-2sin{phi}cos^2{theta}}{sin{theta}}=frac{0}{sin{theta}}=0=RHS$
answered Mar 25 at 17:03
saket kumarsaket kumar
167113
$endgroup$
add a comment |
$begingroup$
$$LHS$$ $=quad2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=frac{2sin{phi}sin^2{theta}-2sin{phi}sin^2{theta}+2sin{phi}cos^2{theta}-2sin{phi}cos^2{theta}}{sin{theta}}=frac{0}{sin{theta}}=0=RHS$
answered Mar 25 at 17:03
saket kumarsaket kumar
167113
$endgroup$
$$LHS$$ $=quad2sin{phi}sin{theta}+frac{2sin{phi}cos{(2theta)}}{sin{theta}}-2cot{theta}sin{phi}cos{theta}=frac{2sin{phi}sin^2{theta}-2sin{phi}sin^2{theta}+2sin{phi}cos^2{theta}-2sin{phi}cos^2{theta}}{sin{theta}}=frac{0}{sin{theta}}=0=RHS$
answered Mar 25 at 17:03
saket kumarsaket kumar
167113
answered Mar 25 at 17:03
saket kumarsaket kumar
167113
answered Mar 25 at 17:03
saket kumarsaket kumar
167113
answered Mar 25 at 17:03
saket kumarsaket kumar
167113
167113
add a comment |
add a comment |
$begingroup$
Hint: It is $$2sin(phi)sin(theta)+frac{2sin(phi)cos(2theta)}{sin(theta)}=frac{2sin(phi)cos(theta)^2}{sin(theta)}$$
answered Mar 25 at 16:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.2k42867
$endgroup$
add a comment |
$begingroup$
Hint: It is $$2sin(phi)sin(theta)+frac{2sin(phi)cos(2theta)}{sin(theta)}=frac{2sin(phi)cos(theta)^2}{sin(theta)}$$
answered Mar 25 at 16:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.2k42867
$endgroup$
add a comment |
$begingroup$
Hint: It is $$2sin(phi)sin(theta)+frac{2sin(phi)cos(2theta)}{sin(theta)}=frac{2sin(phi)cos(theta)^2}{sin(theta)}$$
answered Mar 25 at 16:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.2k42867
$endgroup$
Hint: It is $$2sin(phi)sin(theta)+frac{2sin(phi)cos(2theta)}{sin(theta)}=frac{2sin(phi)cos(theta)^2}{sin(theta)}$$
answered Mar 25 at 16:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.2k42867
answered Mar 25 at 16:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.2k42867
answered Mar 25 at 16:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.2k42867
answered Mar 25 at 16:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.2k42867
79.2k42867
add a comment |
add a comment |
$begingroup$
Multiply by $dfrac{sintheta}{2sinphi}$, and get
$$sin^2{theta}+cos{(2theta)}-cos^2{theta}=0.$$
answered Mar 25 at 16:55
Yves DaoustYves Daoust
133k676232
$endgroup$
add a comment |
$begingroup$
Multiply by $dfrac{sintheta}{2sinphi}$, and get
$$sin^2{theta}+cos{(2theta)}-cos^2{theta}=0.$$
answered Mar 25 at 16:55
Yves DaoustYves Daoust
133k676232
$endgroup$
add a comment |
$begingroup$
Multiply by $dfrac{sintheta}{2sinphi}$, and get
$$sin^2{theta}+cos{(2theta)}-cos^2{theta}=0.$$
answered Mar 25 at 16:55
Yves DaoustYves Daoust
133k676232
$endgroup$
Multiply by $dfrac{sintheta}{2sinphi}$, and get
$$sin^2{theta}+cos{(2theta)}-cos^2{theta}=0.$$
answered Mar 25 at 16:55
Yves DaoustYves Daoust
133k676232
answered Mar 25 at 16:55
Yves DaoustYves Daoust
133k676232
answered Mar 25 at 16:55
Yves DaoustYves Daoust
133k676232
answered Mar 25 at 16:55
Yves DaoustYves Daoust
133k676232
133k676232
add a comment |
add a comment |
1
$begingroup$
I'd start with the non-trigonometric simplification of dividing-through by the common factor of $2sinphi$. (Assuming $sin phineq 0$, of course. But, then, if $sinphi=0$, the equality is trivial.) This reduces a lot of symbolic clutter, clearing the way for some actual trig.
$endgroup$
– Blue
Mar 25 at 16:54