Using Lagranges mean value theorem, show that `sin<>0.`

Answer

Proved

Explanation

Let `f(x)=x-\sin x` and `a` be any positive real number.


(i) `f(x)=x-\sin x` is continuous on `[0, a]`.

(ii) `f^{\prime}(x)=1-\cos x` and this exists uniquely on `(0, a)`.


`\therefore f(x)` is derivable on `(0, \alpha)`.

`\therefore f(x)` satisfies the conditions of Lagrange's Mean value theorem on `[0, a]`.

`\therefore` There exists at least one `c \in(0, a)` such that


`\frac{f(a)-f(0)}{a-0}=f^{\prime}(c)`

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