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# Find Min. value of f(x) = sin6 x + cos6x.  Priyansh Bajaj AskiitiansExpert-IITD
30 Points
10 years ago

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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All the best Man111!!!

Regards,
Priyansh Bajaj

10 years ago

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

10 years ago 