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4sinx+5cosx in the form ksin(x+a)

 Feb 21, 2016

Best Answer 

 #1
avatar+26397 
+30

4sinx+5cosx in the form ksin(x+a)

 

 Asin(x)+Bcos(x)=ksin(x+a)k=A2+B2a=tan1(BA) 

 

Asin(x)+Bcos(x)=ksin(x+a)A=4B=5k=42+52a=tan1(54)k=16+25a=tan1(1.25)k=41a=51.34019174594sin(x)+5cos(x)=41sin(x+51.3401917459)4sin(x)+5cos(x)=6.40312423743sin(x+51.3401917459)

 

laugh

 Feb 22, 2016
 #1
avatar+26397 
+30
Best Answer

4sinx+5cosx in the form ksin(x+a)

 

 Asin(x)+Bcos(x)=ksin(x+a)k=A2+B2a=tan1(BA) 

 

Asin(x)+Bcos(x)=ksin(x+a)A=4B=5k=42+52a=tan1(54)k=16+25a=tan1(1.25)k=41a=51.34019174594sin(x)+5cos(x)=41sin(x+51.3401917459)4sin(x)+5cos(x)=6.40312423743sin(x+51.3401917459)

 

laugh

heureka Feb 22, 2016

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