Answer:
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)
Explanation:
f(x) = Sin x.
f(x+h) = Sin(x+h) = Sin x Cos h + Cos x Sin h
f(x+h) - f(x) = Sin x Cos h - Sin x + Cos x Sin h.
[f(x+h) - f(x)]/h = - Sin x (1 - Cos h)/h + Cos x ×(Sin h/h)
proved
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Verified answer
Answer:
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)
Explanation:
f(x) = Sin x.
f(x+h) = Sin(x+h) = Sin x Cos h + Cos x Sin h
f(x+h) - f(x) = Sin x Cos h - Sin x + Cos x Sin h.
[f(x+h) - f(x)]/h = - Sin x (1 - Cos h)/h + Cos x ×(Sin h/h)
proved