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jagdish singh singh Grade: Upto college level
`        Find Min. value of f(x) = sin6 x + cos6x.`
7 years ago

30 Points
```										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].                          Please  feel free to post as many doubts on our discussion forum as you can. If  you find any questionDifficult to understand - post it here  and we will get you the answer and detailed solution very quickly. We are all  IITians and here to help you in your IIT JEE preparation.All the best  Man111!!!Regards,Askiitians  ExpertsPriyansh Bajaj
```
7 years ago
510 Points
```										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

```
7 years ago
jagdish singh singh
168 Points
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7 years ago
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