Maximum/minimum

f(x)=sin(x-a)/sin(x-b)
Find prove that f(x) assumes maximum or minimum values for a-b=nÏ€ ,n Îµ I

2 Answers

f`(x)=Sin(x-b)Cos(x-a)-Sin(x-a)Cos(x-b)/Sin²(x-b)
f`(x) = Sin(x-b-x+a)/Sin²(x-b)
f`(x)=0
Sin(a-b) = 0
Therefore a-b has to be nâˆ© for n Îµ I

Hence proved

Thank you :-)

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