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De gulden snede
f(x)=sin(x)(sin(x)+2cos(x))f(x)=sin2(x)+2sin(x)cos(x)f′(x)=2sin(x)cos(x)+4cos2(x)−2f′(x)=4cos2(x)+2sin(x)cos(x)−2f′(x)=0}⇒4cos2(x)+2sin(x)cos(x)−2=02cos(2x)+2+sin(2x)−2=02cos(2x)+sin(2x)=02+tan(2x)=0tan(2x)=−2x=38π−12arctan(13)f(38π−12arctan(13))=12+12√5=φ