Find Min. value of f(x) = sin⁶ x + cos⁶x.

Dear Man111,

Solution:- f(x) = (sin² x)³ + (cos² x)³ = (sin² x+ cos²x) [ (sin² x)² + sin² x cos²x +(cos²x)² ]

                         = [ (sin² x + cos²x)² - 2sin² x cos²x + sin² x cos²x ]

                         = [1 - sin² x cos²x]

f(x) is minimum when (sin² x cos²x) is maximum and maximum value of (sin² x cos²x) is 1/2 when x =45⁰

So, min. value of f(x) is 1/2 [ANS].                        

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Priyansh Bajaj

f(x)=sin⁶x   +   cos⁶x

      = (sin²x)³ + (cos²x)    

 f(x)= (sin²x+cos²x)(sin⁴x+cos⁴x-sinxcosx)                                                 a³ + b³ =(a+b)(a² +b² -ab)

f(x)={[(1-cos2x)/2 ]² +[(1+cos2x)/2]² -sin2x/2}                            sin²x=(1-cos2x)/2  ,cos²x=(1+cos2x)/2

f(x)=[1-(sin²2x+sin2x)]/2 .......................1

for f(x) to be minimum ,the subtracting term (sin²2x + sin2x) should be maximum ,

and its maximum value is 2 at x=pi/4,

so minimum value of f(x)=1-2/2=-1/2