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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
 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
 one year ago

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

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

mathsloverBest ResponseYou've already chosen the best response.1
What 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 : \(\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi) + i \sin (\theta + \phi +\Psi)]\)
 one year ago

mathsloverBest 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)
 one year ago

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

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

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

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

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

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

hartnnBest ResponseYou've already chosen the best response.1
i 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.....
 one year ago

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

mathsloverBest 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
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
a= 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 ?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.1
1/a = cos x  i sin x / (cos^2 x  i sin ^2 x)
 one year ago

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

mathsloverBest ResponseYou've already chosen the best response.1
I 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
 one year ago

harsh314Best ResponseYou've already chosen the best response.0
but i get the answer as\[\frac{ 9 }{ \sqrt{2} }\] for (a)
 one year ago

harsh314Best ResponseYou've already chosen the best response.0
at the end i got the required part as \[18(e ^{i (\alpha+ \beta+\gamma)}+e ^{i(\alpha+\beta+\gamma)})\]
 one year ago

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

harsh314Best ResponseYou've already chosen the best response.0
now solving it i \[18\cos(\alpha+\beta +\gamma)/2\]got \[\alpha +\beta +\gamma=\frac{ \pi }{ 4 }\] and hence the required answer
 one year ago

harsh314Best ResponseYou've already chosen the best response.0
\[e ^{i \times \theta}+e ^{i \times \theta}=2 \cos(\theta)\]
 one year ago

harsh314Best ResponseYou've already chosen the best response.0
it is \[18 \frac{ \cos(\alpha +\beta+\gamma) }{ 2 }\]
 one year ago

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

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

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

harsh314Best ResponseYou've already chosen the best response.0
yes the last portion is same what is the value of\[\alpha +\beta +\gamma\]
 one year ago

harsh314Best ResponseYou've already chosen the best response.0
i did it by using\[e ^{i \theta}+e ^{i \theta}=2\cos(\theta),............e ^{i \theta}e ^{i \theta}=2i \sin \theta \]
 one year ago

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

harsh314Best ResponseYou've already chosen the best response.0
see 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
 one year ago

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

harsh314Best ResponseYou've already chosen the best response.0
from the two given equations the two equations can be derived
 one year ago

mathsloverBest ResponseYou've already chosen the best response.1
right 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 : \[\large{(\cos \alpha)^3 + (2\cos \ beta )^3 + (3\cos \gamma)^3 = 3(\cos \alpha )(2\cos \beta) (3\cos \gamma)}\] \[\large{\cos 3 \alpha + 8 \cos 3 \beta +27 \cos 3 \gamma = 18 \cos \alpha \cos \beta \cos \gamma }\]
 one year ago

harsh314Best ResponseYou've already chosen the best response.0
no how did you get so
 one year ago

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

harsh314Best ResponseYou've already chosen the best response.0
get the value of alpha +beta+gamma
 one year ago
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