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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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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
Oh, I see. Is that the final answer?
I mean, is that it? Usually these questions take up more space. >_<
I'm pretty sure it is, yes
you can also use the identity that \[\bf \large \boxed{\sin 2A= 2\sin A \times \cos A }\]
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treating \[4u=2 \times (2u) \]
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