mathslover
  • mathslover
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) \(\sin (\phi + \psi) + 2\sin (\Psi + \theta) + 3 \sin (\theta + \psi)\)
Mathematics
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathslover
  • mathslover
@harsh314 @AravindG @blues @hartnn @Callisto @ghazi @jim_thompson5910
anonymous
  • anonymous
this question is solved, all you need is the sum of angles identity
mathslover
  • mathslover
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 : \(\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi) + i \sin (\theta + \phi +\Psi)]\)

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mathslover
  • mathslover
\[\large{\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi) ]}\] ^ this is what I got for part a) .. I am having problem in part b)
mathslover
  • mathslover
@dumbcow
mathslover
  • mathslover
Got to go now but please answer this question if possible ... Thanks!
mathslover
  • mathslover
Any one who can help me ??
mathslover
  • mathslover
@dumbcow @UnkleRhaukus @hartnn @ghazi @ghass1978 @Callisto @jim_thompson5910 -- Any help or clue ?
mathslover
  • mathslover
I am getting this when I tried Part 2) : \[\large{\sin 3\theta + 8\sin 3 \phi + 27 \sin 3 \psi = 18 \sin ( \theta + \phi + \psi )}\]
mathslover
  • mathslover
OK leave it can any one prove that if : a + 2b + 3c = 0 then : \[\large{\frac{1}{2} + \frac{2}{b} + \frac{3}{c} = 0 }\]
anonymous
  • anonymous
the answer is SAM IS AMAZINGLY BEAUTIFUL no need to thank me ;D
hartnn
  • hartnn
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 (x-y)] = 4 c (x-y) = 1 x-y = 0 x=y. similarly if i take , sx + 3sz = -2sy ----->(1) cx + 3cz = -2cy----->(2) then u get x-z = \(\pi\) so, i have now, \(\theta = \phi = \pi +\psi \) substitute this in b, and continue.....
hartnn
  • hartnn
not sure whether that would be useful...
mathslover
  • mathslover
@hartnn can you help me in proving : 1/a + 2/b + 3/c = 0 if a + 2b + 3c = 0
mathslover
  • mathslover
@shubhamsrg
shubhamsrg
  • shubhamsrg
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 ?
mathslover
  • mathslover
1/a = cos x - i sin x / (cos^2 x - i sin ^2 x)
experimentX
  • experimentX
i^2 = -1
mathslover
  • mathslover
oh yep...
mathslover
  • mathslover
i got it now ... thanks!
mathslover
  • mathslover
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
anonymous
  • anonymous
but i get the answer as\[\frac{ 9 }{ \sqrt{2} }\] for (a)
mathslover
  • mathslover
how?
anonymous
  • anonymous
at the end i got the required part as \[18(e ^{i (\alpha+ \beta+\gamma)}+e ^{-i(\alpha+\beta+\gamma)})\]
mathslover
  • mathslover
yep than equate the real parts
anonymous
  • anonymous
now solving it i \[18\cos(\alpha+\beta +\gamma)/2\]got \[\alpha +\beta +\gamma=\frac{ \pi }{ 4 }\] and hence the required answer
anonymous
  • anonymous
\[e ^{i \times \theta}+e ^{-i \times \theta}=2 \cos(\theta)\]
anonymous
  • anonymous
it is \[18 \frac{ \cos(\alpha +\beta+\gamma) }{ 2 }\]
mathslover
  • mathslover
I think we have : \[{\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi + i [\sin 3 \theta + \sin 3\phi +\\ \sin 3 \Psi ] = 18[\cos (\theta + \phi +\Psi) + i \sin (\theta + \phi +\Psi)] }\]
mathslover
  • mathslover
and then equating the real parts we get : \[\large{\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi)]}\]
mathslover
  • mathslover
Which mthod did you follow ? mine (mentioned on the top ) ? or any other ... ?
anonymous
  • anonymous
yes the last portion is same what is the value of\[\alpha +\beta +\gamma\]
anonymous
  • anonymous
i did it by using\[e ^{i \theta}+e ^{-i \theta}=2\cos(\theta),............e ^{i \theta}-e ^{-i \theta}=2i \sin \theta \]
mathslover
  • mathslover
Well so how you got to this : \[\large{18 [e^{ i ( \theta + \phi + \psi)} + e^{-i (\theta + \phi +\psi)}]}\]?
anonymous
  • anonymous
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
mathslover
  • mathslover
you added the both equations ? : \[\large{e^{- i\alpha} + e^{i \alpha} + 2e^{-i \beta} + 2e^{i \beta} + 3e^{-i\gamma} + 3e^{i \gamma} = 0 }\] \[\large{2 ( \cos \alpha + 2 \cos \beta + 3 \cos \gamma )= 0}\]
anonymous
  • anonymous
from the two given equations the two equations can be derived
mathslover
  • mathslover
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 : \[\large{(\cos \alpha)^3 + (2\cos \ beta )^3 + (3\cos \gamma)^3 = 3(\cos \alpha )(2\cos \beta) (3\cos \gamma)}\] \[\large{\cos 3 \alpha + 8 \cos 3 \beta +27 \cos 3 \gamma = 18 \cos \alpha \cos \beta \cos \gamma }\]
anonymous
  • anonymous
no how did you get so
mathslover
  • mathslover
I think I did wrong somewhere
anonymous
  • anonymous
get the value of alpha +beta+gamma

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