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mathslover

  • 3 years ago

If sinθ+2sinϕ+3sinΨ=0 and cosθ+2cosϕ+3cosΨ=0 ; then evaluate : a) cos3θ+8cos3ϕ+27cos3Ψ b) sin(ϕ+ψ)+2sin(Ψ+θ)+3sin(θ+ψ)

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  1. mathslover
    • 3 years ago
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    I did the first one ... here : http://openstudy.com/study#/updates/511c70c9e4b06821731b25bc but second one is hard for me... can any1 help me?

  2. .Sam.
    • 3 years ago
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    For 2nd one did you try to use the tirg identities? e.g. sin(A+B)=sinAcosB+cosAsinB

  3. mathslover
    • 3 years ago
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    hmn... no but ok i will try it now..

  4. mathslover
    • 3 years ago
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    But where and how to use that formula @.Sam.

  5. mathslover
    • 3 years ago
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    OK leave it can any one prove that if : a + 2b + 3c = 0 then : \[\large{\frac{1}{2} + \frac{2}{b} + \frac{3}{c} = 0 }\]

  6. .Sam.
    • 3 years ago
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    For \[3 \sin (\theta +\psi )+2 \sin (\theta +\Psi )+\sin (\psi +\phi )\] You just use it for each term You'll get \[\small 3 \sin(\theta )\cos (\psi )+3\cos (\theta ) \sin (\psi )+2\sin (\theta ) \cos(\Psi )+2 \cos(\theta ) \sin(\Psi )+\cos(\psi ) \sin(\phi )+ \sin(\psi ) \cos (\phi )\] Then try to factor it and apply ( sinθ+2sinϕ+3sinΨ=0 and cosθ+2cosϕ+3cosΨ=0 )

  7. mathslover
    • 3 years ago
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    But is that proving or just verifying ?

  8. .Sam.
    • 3 years ago
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    its actually simplifying, try to factor that then your factored equation can apply this ( sinθ+2sinϕ+3sinΨ=0 and cosθ+2cosϕ+3cosΨ=0 ), which makes the equation smaller by equating them to 0.

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