anonymous
  • anonymous
Verify the identity. sin (4u) = 2 sin (2u) cos (2u) Please help, verifying these kind of equations is tricky for me.
Mathematics
schrodinger
  • schrodinger
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slaaibak
  • slaaibak
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
anonymous
  • anonymous
Oh, I see. Is that the final answer?
anonymous
  • anonymous
I mean, is that it? Usually these questions take up more space. >_<

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slaaibak
  • slaaibak
I'm pretty sure it is, yes
AravindG
  • AravindG
you can also use the identity that \[\bf \large \boxed{\sin 2A= 2\sin A \times \cos A }\]
AravindG
  • AravindG
treating \[4u=2 \times (2u) \]

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