Dreew
Dreew
13-05-2014
Mathematics
contestada
how do you do...
cos2x-sin2x=1-2sin2x
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kate200468
kate200468
13-05-2014
[tex]sin (2 \alpha) =2\cdot sin \alpha \cdot cos \alpha \\cos(2 \alpha )=1-2sin^2 \alpha \\-------------------- \\cos2x-sin2x=1-2sin2x\\cos2x+sin2x=1\\1-2sin^2x+2sinx\cdot cosx=1\\-2sinx(sin x-cosx)=0\\ -2sinx=0\ \ \ \ or\ \ \ \ sinx-cosx=0\\\\1)\ -2sinx=0\ \ \Leftrightarrow\ \ sinx=0\ \ \Leftrightarrow\ \ x_1=k \pi \ \ and\ \ k\in I\\2)\ \ sinx-cosx=0\ \ \Leftrightarrow\ \ sinx=cosx\ /:cosx\ \ \ \ \ \wedge\ \ \ \ cos x \neq 0\\tanx=1\ \ \Leftrightarrow\ \ x_2= \frac{ \pi }{4} +k \pi \ \ and\ \ k\in I[/tex]
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