using laplace transform table and convolution theorem show that L{sin(ax+b)}=s sin(b)+a cos(b)/s二次方+a二次方 a, b are constants
please steps by steps !
sin(ax+b)=sin(ax)cos(b)+cos(ax)sin(b)
L{sin(ax)}= a/(s^2+a^2)
L{cos(ax)}= s/(s^2+a^2)
so that,
L{sin(ax+b)}
=L{sin(ax)cos(b)+cos(ax)sin(b)}
=sin(b)L{cos(ax)}+ cos(b)L{sin(ax)}
= sin(b)*[s/(s^2+a^2)]+ cos(b)*[ a/(s^2+a^2) ]
= [ s sin(b) + a cos(b)]/(s^2+a^2)
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Verified answer
sin(ax+b)=sin(ax)cos(b)+cos(ax)sin(b)
L{sin(ax)}= a/(s^2+a^2)
L{cos(ax)}= s/(s^2+a^2)
so that,
L{sin(ax+b)}
=L{sin(ax)cos(b)+cos(ax)sin(b)}
=sin(b)L{cos(ax)}+ cos(b)L{sin(ax)}
= sin(b)*[s/(s^2+a^2)]+ cos(b)*[ a/(s^2+a^2) ]
= [ s sin(b) + a cos(b)]/(s^2+a^2)