Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Find Min. value of f(x) = sin 6 x + cos 6 x.

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

Grade:Upto college level

3 Answers

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].                        

Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult 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 Experts
Priyansh Bajaj

vikas askiitian expert
509 Points
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

 

 

jagdish singh singh
173 Points
10 years ago

2346_24699_gif.latex (2).gif

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free