4sinx+5cosx in the form ksin(x+a)
A⋅sin(x)+B⋅cos(x)=k⋅sin(x+a)k=√A2+B2a=tan−1(BA)
A⋅sin(x)+B⋅cos(x)=k⋅sin(x+a)A=4B=5k=√42+52a=tan−1(54)k=√16+25a=tan−1(1.25)k=√41a=51.3401917459∘4⋅sin(x)+5⋅cos(x)=√41⋅sin(x+51.3401917459∘)4⋅sin(x)+5⋅cos(x)=6.40312423743⋅sin(x+51.3401917459∘)
4sinx+5cosx in the form ksin(x+a)
A⋅sin(x)+B⋅cos(x)=k⋅sin(x+a)k=√A2+B2a=tan−1(BA)
A⋅sin(x)+B⋅cos(x)=k⋅sin(x+a)A=4B=5k=√42+52a=tan−1(54)k=√16+25a=tan−1(1.25)k=√41a=51.3401917459∘4⋅sin(x)+5⋅cos(x)=√41⋅sin(x+51.3401917459∘)4⋅sin(x)+5⋅cos(x)=6.40312423743⋅sin(x+51.3401917459∘)