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anonymous

  • one year ago

Solve sin A + sin 2A + sin 3A + sin 4A = 0, for 0 < A < 180 (0 and 180 inclusive)

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  1. anonymous
    • one year ago
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    I've tried using factor formula but still did not manage to get the answer, not sure if factor formula is the right method

  2. anonymous
    • one year ago
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    I rearrange to sin 4A + sin 2A + sin 3A + sin A = 0, and after applying factor formula, 2 sin 3A cos A + 2 sin 2A cos A = 0 2 cos A ( sin 3A + sin 2A) = 0 2 cos A ( sin 5/2 A cos 1/2 A) = 0

  3. anonymous
    • one year ago
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    @thomaster @Preetha

  4. mathmath333
    • one year ago
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    \(A=0^{\circ}\) is one solution by obsrvation

  5. welshfella
    • one year ago
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    @HatcrewS you are on the right road to solving it

  6. anonymous
    • one year ago
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    @welshfella how do I continue from there ? T.T

  7. welshfella
    • one year ago
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    i think there is an error in your last line

  8. welshfella
    • one year ago
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    2 cos A ( sin 5/2 A cos 1/2 A) = 0 should be 2 cos A ( 2 sin 5/2 A cos 1/2 A) = 0 now its just a matter of equating the 3 terms to 0

  9. anonymous
    • one year ago
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    @welshfella oh okay i'll try again thanks!

  10. welshfella
    • one year ago
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    2 cos A = 0 A = 90 degrees

  11. anonymous
    • one year ago
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    I solve and got 0, 90, 144, 180 but im still missing 72(answer sheet shows that 72 is one of the answer). Can someone help me get 72 ?

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