Im doing derivatives I understand them but my book is confusing the hell out of me.
this what my book is showing
f(x) =2sinx-cos2x original function
f"(x)=2cosx+2sin2x=0 btw we are doing extrema thats why the derivative is set to 0
then is says"sin2x=2cosxsinx"
wtf since when? then it goes on simplifying and factoring
2cosx+4cosxsinx
2(cosx)(1+2sinx)=0
can someone please explain to me how sin2x is equal to 2cosxsinx
calculus how does sin2x=2cosx*sinx?
2011-03-23 6:00 am
回答 (4)
2016-11-16 9:30 pm
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2011-03-23 6:05 am
Sin(A+B) = sinAcosB+sinBcos A , if A=B
sin2A= 2sinAcosA
sin2A= 2sinAcosA
2011-03-23 6:03 am
sin(2x) = 2sinxcosx is a double angle identity that you should have learned in trig.
2011-03-23 9:06 am
Rather a long story !
sin ( A + B) = sin A cos B + cos A sin B
sin 2Ө = sin ( Ө + Ө )
sin ( Ө + Ө ) = sin Ө cos Ө + cos Ө sin Ө
sin ( Ө + Ө ) = sin Ө cos Ө + sin Ө cos Ө
sin ( Ө + Ө ) = 2 sin Ө cos Ө
sin ( A + B) = sin A cos B + cos A sin B
sin 2Ө = sin ( Ө + Ө )
sin ( Ө + Ө ) = sin Ө cos Ө + cos Ө sin Ө
sin ( Ө + Ө ) = sin Ө cos Ө + sin Ө cos Ө
sin ( Ө + Ө ) = 2 sin Ө cos Ө
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