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

1. mathslover

@harsh314 @AravindG @blues @hartnn @Callisto @ghazi @jim_thompson5910

2. mathsmind

this question is solved, all you need is the sum of angles identity

3. 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 : $$\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi) + i \sin (\theta + \phi +\Psi)]$$

4. 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)

5. mathslover

@dumbcow

6. mathslover

Got to go now but please answer this question if possible ... Thanks!

7. mathslover

Any one who can help me ??

8. mathslover

@dumbcow @UnkleRhaukus @hartnn @ghazi @ghass1978 @Callisto @jim_thompson5910 -- Any help or clue ?

9. mathslover

I am getting this when I tried Part 2) : $\large{\sin 3\theta + 8\sin 3 \phi + 27 \sin 3 \psi = 18 \sin ( \theta + \phi + \psi )}$

10. mathslover

OK leave it can any one prove that if : a + 2b + 3c = 0 then : $\large{\frac{1}{2} + \frac{2}{b} + \frac{3}{c} = 0 }$

11. SamGracie

the answer is SAM IS AMAZINGLY BEAUTIFUL no need to thank me ;D

12. 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.....

13. hartnn

not sure whether that would be useful...

14. mathslover

@hartnn can you help me in proving : 1/a + 2/b + 3/c = 0 if a + 2b + 3c = 0

15. mathslover

@shubhamsrg

16. 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 ?

17. mathslover

1/a = cos x - i sin x / (cos^2 x - i sin ^2 x)

18. experimentX

i^2 = -1

19. mathslover

oh yep...

20. mathslover

i got it now ... thanks!

21. 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

22. harsh314

but i get the answer as$\frac{ 9 }{ \sqrt{2} }$ for (a)

23. mathslover

how?

24. harsh314

at the end i got the required part as $18(e ^{i (\alpha+ \beta+\gamma)}+e ^{-i(\alpha+\beta+\gamma)})$

25. mathslover

yep than equate the real parts

26. harsh314

now solving it i $18\cos(\alpha+\beta +\gamma)/2$got $\alpha +\beta +\gamma=\frac{ \pi }{ 4 }$ and hence the required answer

27. harsh314

$e ^{i \times \theta}+e ^{-i \times \theta}=2 \cos(\theta)$

28. harsh314

it is $18 \frac{ \cos(\alpha +\beta+\gamma) }{ 2 }$

29. mathslover

I think we have : ${\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi + i [\sin 3 \theta + \sin 3\phi +\\ \sin 3 \Psi ] = 18[\cos (\theta + \phi +\Psi) + i \sin (\theta + \phi +\Psi)] }$

30. mathslover

and then equating the real parts we get : $\large{\cos 3\theta + 8\cos 3 \phi + 27\cos 3 \Psi = 18[\cos (\theta + \phi +\Psi)]}$

31. mathslover

Which mthod did you follow ? mine (mentioned on the top ) ? or any other ... ?

32. harsh314

yes the last portion is same what is the value of$\alpha +\beta +\gamma$

33. harsh314

i did it by using$e ^{i \theta}+e ^{-i \theta}=2\cos(\theta),............e ^{i \theta}-e ^{-i \theta}=2i \sin \theta$

34. mathslover

Well so how you got to this : $\large{18 [e^{ i ( \theta + \phi + \psi)} + e^{-i (\theta + \phi +\psi)}]}$?

35. harsh314

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

36. mathslover

you added the both equations ? : $\large{e^{- i\alpha} + e^{i \alpha} + 2e^{-i \beta} + 2e^{i \beta} + 3e^{-i\gamma} + 3e^{i \gamma} = 0 }$ $\large{2 ( \cos \alpha + 2 \cos \beta + 3 \cos \gamma )= 0}$

37. harsh314

from the two given equations the two equations can be derived

38. 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 : $\large{(\cos \alpha)^3 + (2\cos \ beta )^3 + (3\cos \gamma)^3 = 3(\cos \alpha )(2\cos \beta) (3\cos \gamma)}$ $\large{\cos 3 \alpha + 8 \cos 3 \beta +27 \cos 3 \gamma = 18 \cos \alpha \cos \beta \cos \gamma }$

39. harsh314

no how did you get so

40. mathslover

I think I did wrong somewhere

41. harsh314

get the value of alpha +beta+gamma