A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 one year ago
If \(\sin \theta + 2\sin \phi + 3\sin \Psi = 0 \) and \(\cos \theta + 2\cos \phi + 3\cos \Psi = 0\) ; then evaluate :
a) \(\cos 3\theta + 8\cos 3 \phi + 27 \cos 3\Psi \)
b) \(\sin (\phi + \psi) + 2\sin (\Psi + \theta) + 3 \sin (\theta + \psi)\)
 one year ago
If \(\sin \theta + 2\sin \phi + 3\sin \Psi = 0 \) and \(\cos \theta + 2\cos \phi + 3\cos \Psi = 0\) ; then evaluate : a) \(\cos 3\theta + 8\cos 3 \phi + 27 \cos 3\Psi \) b) \(\sin (\phi + \psi) + 2\sin (\Psi + \theta) + 3 \sin (\theta + \psi)\)

This Question is Closed

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1@harsh314 @AravindG @blues @hartnn @Callisto @ghazi @jim_thompson5910

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.0this question is solved, all you need is the sum of angles identity

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1What I tried is : I did it for Part a) .. which is as follows Let \(\sin \theta + 2\sin \phi + 3\sin \Psi =0\) 1) and \(\cos \theta + 2\cos \phi + 3\cos \Psi = 0\) be equation 2) Multiply equn 1) by iota (\(i\) ) ... and then add eqn 1) and 2) : we get : \(\cos \theta + i \sin \theta + 2\cos \phi + 2i\sin\phi + 3\cos \Psi + 3i \cos \Psi = 0\) Say \(\cos \theta + i \sin \theta =a \) , \(\cos \phi + i\sin \phi =b\) ,\(\cos \Psi + i \sin \Psi = c\) We get a + 2b + 3c = 0 then we get : \(a^3 + 8b^3 + 27c^3 = 18abc\) then simplifying it and equating the real parts I got : \(\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi) + i \sin (\theta + \phi +\Psi)]\)

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1\[\large{\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi) ]}\] ^ this is what I got for part a) .. I am having problem in part b)

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Got to go now but please answer this question if possible ... Thanks!

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Any one who can help me ??

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1@dumbcow @UnkleRhaukus @hartnn @ghazi @ghass1978 @Callisto @jim_thompson5910  Any help or clue ?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1I am getting this when I tried Part 2) : \[\large{\sin 3\theta + 8\sin 3 \phi + 27 \sin 3 \psi = 18 \sin ( \theta + \phi + \psi )}\]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1OK leave it can any one prove that if : a + 2b + 3c = 0 then : \[\large{\frac{1}{2} + \frac{2}{b} + \frac{3}{c} = 0 }\]

SamGracie
 one year ago
Best ResponseYou've already chosen the best response.0the answer is SAM IS AMAZINGLY BEAUTIFUL no need to thank me ;D

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1i think i've got something. To simplify writing, i'll take \(\theta =x , \phi =y, \psi =z, sin =s, cos =c.\) sx + 2sy = 3sz >(1) cx +2cy = 3cz>(2) Square both (1) and (2) and then add. 1+ 4 [sxsy + cxcy ] + 4 = 9 4 [c (xy)] = 4 c (xy) = 1 xy = 0 x=y. similarly if i take , sx + 3sz = 2sy >(1) cx + 3cz = 2cy>(2) then u get xz = \(\pi\) so, i have now, \(\theta = \phi = \pi +\psi \) substitute this in b, and continue.....

hartnn
 one year ago
Best ResponseYou've already chosen the best response.1not sure whether that would be useful...

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1@hartnn can you help me in proving : 1/a + 2/b + 3/c = 0 if a + 2b + 3c = 0

shubhamsrg
 one year ago
Best ResponseYou've already chosen the best response.0a= cosx + isinx b= cosy + isiny c= cosz +isinz =>1/a = cosx  isinx 1/b= cosy isiny 1/c = cosz isinz Substitute and see if it helps ?

mathslover
 one year ago
Best ResponseYou've already chosen the best response.11/a = cos x  i sin x / (cos^2 x  i sin ^2 x)

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1i got it now ... thanks!

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1I did like this : \[\large{a + 2b + 3c = 0}\] \[\large{ \bar {\bar{a}+2b+3\bar{c}} = 0}\] Which is : \[\large{e^{i \theta} + 2e^{i \phi} + 3e^{i \Psi} = 0}\] Which I can write as : \[\large{\frac{1}{a} + \frac{2}{b} + \frac{3}{c} = 0}\] and then I get : \[\large{bc + 2ac + 3ab = 0}\] By simplifying this I get the required answer .... i.e. 0

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0but i get the answer as\[\frac{ 9 }{ \sqrt{2} }\] for (a)

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0at the end i got the required part as \[18(e ^{i (\alpha+ \beta+\gamma)}+e ^{i(\alpha+\beta+\gamma)})\]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1yep than equate the real parts

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0now solving it i \[18\cos(\alpha+\beta +\gamma)/2\]got \[\alpha +\beta +\gamma=\frac{ \pi }{ 4 }\] and hence the required answer

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0\[e ^{i \times \theta}+e ^{i \times \theta}=2 \cos(\theta)\]

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0it is \[18 \frac{ \cos(\alpha +\beta+\gamma) }{ 2 }\]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1I think we have : \[{\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi + i [\sin 3 \theta + \sin 3\phi +\\ \sin 3 \Psi ] = 18[\cos (\theta + \phi +\Psi) + i \sin (\theta + \phi +\Psi)] }\]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1and then equating the real parts we get : \[\large{\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi)]}\]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Which mthod did you follow ? mine (mentioned on the top ) ? or any other ... ?

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0yes the last portion is same what is the value of\[\alpha +\beta +\gamma\]

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0i did it by using\[e ^{i \theta}+e ^{i \theta}=2\cos(\theta),............e ^{i \theta}e ^{i \theta}=2i \sin \theta \]

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1Well so how you got to this : \[\large{18 [e^{ i ( \theta + \phi + \psi)} + e^{i (\theta + \phi +\psi)}]}\]?

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0see the equations would be \[e ^{i \alpha}+ 2e^{i \beta}+3 e ^{i \gamma}=0\] and \[e ^{i \alpha}+ 2e^{i \beta}+3 e ^{i \gamma}=0\]and from these we get \[e ^{2i \gamma}=e ^{i (\alpha+\beta)}\] cubing the first 2 equations and using the third one we get the mentioned thing

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1you added the both equations ? : \[\large{e^{ i\alpha} + e^{i \alpha} + 2e^{i \beta} + 2e^{i \beta} + 3e^{i\gamma} + 3e^{i \gamma} = 0 }\] \[\large{2 ( \cos \alpha + 2 \cos \beta + 3 \cos \gamma )= 0}\]

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0from the two given equations the two equations can be derived

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1right so did I do right above .. ? and then : \[\large{ 2( \cos \alpha + 2\cos \beta + 3\cos \gamma )= 0}\] \[\large{ \cos \alpha + 2\cos \beta + 3 \cos \gamma = 0}\] Now since \(\cos \alpha + 2\ cos \beta + 3\ cos \gamma =0\) therefore : \[\large{(\cos \alpha)^3 + (2\cos \ beta )^3 + (3\cos \gamma)^3 = 3(\cos \alpha )(2\cos \beta) (3\cos \gamma)}\] \[\large{\cos 3 \alpha + 8 \cos 3 \beta +27 \cos 3 \gamma = 18 \cos \alpha \cos \beta \cos \gamma }\]

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0no how did you get so

mathslover
 one year ago
Best ResponseYou've already chosen the best response.1I think I did wrong somewhere

harsh314
 one year ago
Best ResponseYou've already chosen the best response.0get the value of alpha +beta+gamma
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.