given that f(x) = sin(x), show that f(x+h)-f(x)/h = sinx(cos(h)-1/h) + cos(x)(sin(h)/h)

Question

given that f(x) = sin(x), show that f(x+h)-f(x)/h = sinx(cos(h)-1/h) + cos(x)(sin(h)/h) precal question

 

Answer 1

Difference quotient:     [ƒ(x + h) - ƒ(x)]/h

Given:     ƒ(x) = sin(x)     ==>     ƒ(x + h) = sin(x + h)

Recall sum/difference identity of sin function:

     sin(θ ± φ) = sin(θ)cos(φ) ± cos(θ)sin(φ)

Therefore,

        ƒ(x + h) = sin(x + h) = sin(x)cos(h) + cos(x)sin(h)

Thus,

     [ƒ(x + h) - ƒ(x)]/h = [sin(x + h) - sin(x)]/h

                                = [(sin(x)cos(h) + cos(x)sin(h)) - sin(x)]/h

                                = [(sin(x)cos(h) - sin(x) + cos(x)sin(h))]/h

                                = [(sin(x)(cos(h) - 1)) + cos(x)sin(h)]/h

                                = (sin(x)(cos(h) - 1))/h + (cos(x)sin(h))/h

                                = sin(x)((cos(h) - 1)/h) + cos(x)(sin(h)/h)

Answer 2

[f(x+h)-f(x)]/h

= [sin(x+h) - sin(x)]/h

= [sin(x)cos(h) + cos(x)sin(h) - sin(x)]/h

= sin(x)(cos(h) - 1)/h + cos(x)sin(h)/h

QED