Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

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
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

this question is solved, all you need is the sum of angles identity
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 : \(\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi) + i \sin (\theta + \phi +\Psi)]\)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question