How do I show the following...
2sinx sin2x = 4cosx sin^2(x)
2sinx sin2x = 4cosx sin^2(x)
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2sinx . sin2x
We have this formula : sin2x = 2sinxcosx
=> 2sinx . 2sinxcosx
= 4sin²xcosx
We have this formula : sin2x = 2sinxcosx
=> 2sinx . 2sinxcosx
= 4sin²xcosx
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2sinx sin2x = 4cosx sin^2(x)
2sinx(2sinx cosx)
4sin^2x cosx = 4cosx sin^2(x)
just rearange since it would all equal the same answer
and sin2x naturally is equal to (2sinx cosx) for some reason
2sinx(2sinx cosx)
4sin^2x cosx = 4cosx sin^2(x)
just rearange since it would all equal the same answer
and sin2x naturally is equal to (2sinx cosx) for some reason