How do you determine the amplitude of displaystyley=5sinx and displaystyley=3cosx ?
Alan P. Apr 25, 2015 Bothdisplaystylesinleft(x ight) ext and cosleft(x ight)have a Range ofdisplaystyleleft<-1,+1 ight>displaystyley=5sinleft(x ight) ...

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displaystylefracleft.dy ight.left.dx ight.=cosx Explanation: displaystylefracdleft.dx ight.left( extconstant ight)=0displaystylefracdleft.dx ight.left(sinx ight)=cosx ...
Anuj Baskota Oct 23, 2015 Here, Since it is sin2x, for its periodicityx= 360or,2x= 360So, x= 180which means that this graph completes one cycle in 180 degrees or 1pie.Now the ...
Amplitude :displaystyle5 Period :displaystyle1 Explanation:A little explanation would be quite adequate for this problem.To determine the Amplitude, put the smallest and largest ...
displaystylefracleft.dy ight.left.dx ight.=fracycosleft(xy ight)1-xcosleft(xy ight) Explanation:Assuming you are differentiating ...

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Edit: Completely rewritten to correct a misreading of the original problem. You know that the graphs of y= an x and y=2sin x cross at x=0. Where else do they cross? You need to solve 2sin x= an x=fracsin xcos x;, ...
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left< eginarray l l 2 & 3 \ 5 & 4 endarray ight> left< eginarray l l l 2 & 0 & 3 \ -1 & 1 & 5 endarray ight>

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