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Trigonometric Inequalities

Problem: Show that cos(sinx)>sin(cosx) for each real x. Sol. Consider the equation Taking + we get, which is not possible Taking – we get, which is again not possible Hence the equation has no solution. Now consider the function . The function is continuous on , therefore exactly one of the statements or for all is […]

Problem: Show that cos(sinx)>sin(cosx) for each real x.

Sol. Consider the equation cos(sinx)=sin(cosx)⇒cos(sinx)=cos(π2−cosx) ⇒cos(sinx)=cos(π2−cosx) ⇒sinx=2nπ±(π2−cosx)
Taking + we get, sinx+cosx=2nπ+π2,n∈Z which is not possible

Taking – we get, sinx−cosx=2nπ+π2,n∈Z which is again not possible

Hence the equation cos(sinx)=sin(cosx) has no solution.

Now consider the function f(x)=cos(sinx)−sin(cosx). The function f is continuous on R , therefore

exactly one of the statements f(x)>0 or f(x)<0 for all x∈R is true beacause if it is not so there would exist two real numbers a and b (say) such that

f(a)f(b)<0 which would imply that the function f has a zero between a and b by Interemediate value property of continuous function, which is a contradiction. Further f(0)=1−sin(1)>0,

Hence cos(sinx)>sin(cosx) for all x∈R

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