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mathslover
 3 years 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) \(\sin (\phi + \psi) + 2\sin (\Psi + \theta) + 3 \sin (\theta + \psi)\)
mathslover
 3 years 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) \(\sin (\phi + \psi) + 2\sin (\Psi + \theta) + 3 \sin (\theta + \psi)\)

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

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

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

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

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

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

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

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

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

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

mathslover
 3 years 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

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

anonymous
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1yep than equate the real parts

anonymous
 3 years 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

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

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

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

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0from the two given equations the two equations can be derived

mathslover
 3 years 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 : \[\large{(\cos \alpha)^3 + (2\cos \ beta )^3 + (3\cos \gamma)^3 = 3(\cos \alpha )(2\cos \beta) (3\cos \gamma)}\] \[\large{\cos 3 \alpha + 8 \cos 3 \beta +27 \cos 3 \gamma = 18 \cos \alpha \cos \beta \cos \gamma }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no how did you get so

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0get the value of alpha +beta+gamma
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