Knowing that sine and cosine are a fundamental set of solutions to y''+y=o, and by assuming d(cos(x))/dx=-sin(x), d(sin(x))/dx=cos(x), sin(0)=0, and cos(0)=1, show that sin(x+u)=sin(x)cos(u)+cos(x)sin(u).

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- March 16th 2010, 10:59 AMscottie22Using y''+y=0 to prove trig identities
Knowing that sine and cosine are a fundamental set of solutions to y''+y=o, and by assuming d(cos(x))/dx=-sin(x), d(sin(x))/dx=cos(x), sin(0)=0, and cos(0)=1, show that sin(x+u)=sin(x)cos(u)+cos(x)sin(u).