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BlingBlong

  • 2 years ago

So I have the function sin(2x) = 2 sin(x) = 2sin(x)cos(x) = 2sin(x) = (cos(x) − 1) = 0 but the text book claims that 2sin(x) (cos(x) − 1) = 0 Can anyone tell me why this is? Did I overlook something?

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  1. TuringTest
    • 2 years ago
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    \[2\sin x(\cos x-1)\neq0\]try it on your calculator

  2. TuringTest
    • 2 years ago
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    the 2 makes no difference so the assertion is that\[\sin x(\cos x-1)=0\]which is just wrong

  3. TuringTest
    • 2 years ago
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    ...as an identity I mean

  4. BlingBlong
    • 2 years ago
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    Furthermore it states that the solution to 2sin(x) is equal to x = 0, π, 2π. Which doesn't make sense to me seeing as it has the integer 2 in front of it

  5. TuringTest
    • 2 years ago
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    \[2\sin x(\cos x-1)=0\to\sin x(\cos x-1)=0\]because 2 cannot be zero, so we have\[\sin x=0\to x=n\pi, n\in\mathbb N\]but you may have figure that along with the cosine part

  6. TuringTest
    • 2 years ago
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    but that's just going on that part of the formula

  7. TuringTest
    • 2 years ago
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    ok I see your problem...

  8. TuringTest
    • 2 years ago
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    \[\sin(2x)=2\sin x\]\[2\sin x\cos x=2\sin x\]\[\sin x\cos x=\sin x\]now you can't just divide by sinx because that assumes that sinx is not zero, which it might be!, so we have to keep it around and factor if we want all the solutions.\[\sin x\cos x-\sin x=0\]\[\sin x(\cos x-1)=0\]so we have to solve\[\sin x=0\]\[\cos x-1=0\]

  9. BlingBlong
    • 2 years ago
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    sorry I forgot to mention that the domain is restricted to [0,2pi]

  10. TuringTest
    • 2 years ago
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    no big deal with the domain, makes it simpler

  11. BlingBlong
    • 2 years ago
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    oh ok I see :) thanks ugh

  12. TuringTest
    • 2 years ago
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    welcome!

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