Answer to Question #197351 in Real Analysis for Abu sufian

Question

Show that sinx and cosx are a bound variation on a finite interval

Answer 1

$\text{We know , if a function is Lipschitz function then it is bounded variation.}\newline \text{Given,} f_1(x)=sinx. \,\\ |f_1(x_1)-f_1(x_2)|\newline =|sin(x_1)-sin(x_2)|\hspace{1cm}(\text{By trigonometry identity}\newline =|2 cos(\frac{x_1+x_2}{2}) sin(\frac{x_1-x_2}{2})|\newline \leq|2|.1.|\frac{x_1-x_2}{2}|\newline =|x_1-x_2|\newline \text{ we get k=1.}\newline \text{ Therefore, the given function is lipschitz function.}\newline \text{This implies, the given function is bounded variation.} \text{Given,} f_2(x)=cosx.\\ |f_1(x_1)-f_1(x_2)|\newline =|cos(x_1)-cos(x_2)|\hspace{1cm}(\text{By trigonometry identity}\newline =|-2 sin(\frac{x_1+x_2}{2}) sin(\frac{x_1-x_2}{2})|\newline \leq|2|.1.|\frac{x_1-x_2}{2}|\newline =|x_1-x_2|\newline \text{ we get k=1.}\newline \text{ Therefore, the given function is lipschitz function.}\newline \text{This implies, the given function is bounded variation.}$