Another Trig Identity to Verify...
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Another Trig Identity to Verify...

[From: ] [author: ] [Date: 11-11-26] [Hit: ]
......
How do I show the following...

2sinx sin2x = 4cosx sin^2(x)

-
2sinx . sin2x

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
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