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6532bunny

  • 2 years ago

Verify the identity. sin (4u) = 2 sin (2u) cos (2u) Please help, verifying these kind of equations is tricky for me.

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  1. slaaibak
    • 2 years ago
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    Hmm. you can try doing it like this: sin(2u + 2u) = sin(2u) * cos(2u) + cos(2u) * sin(2u) = 2sin(2u)*cos(2u) Note: I referred to: sin(A + B) = sinAcosB + cosAsinB

  2. 6532bunny
    • 2 years ago
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    Oh, I see. Is that the final answer?

  3. 6532bunny
    • 2 years ago
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    I mean, is that it? Usually these questions take up more space. >_<

  4. slaaibak
    • 2 years ago
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    I'm pretty sure it is, yes

  5. AravindG
    • 2 years ago
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    you can also use the identity that \[\bf \large \boxed{\sin 2A= 2\sin A \times \cos A }\]

  6. AravindG
    • 2 years ago
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    treating \[4u=2 \times (2u) \]

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