If m= sin6x + cos6x, then m belongs to the interval


(a) [7/8, 5/4]  (b) [1/2 , 5/8]  (c) [1/4,  1]


Explain pls...

2 years ago

Share

Answers : (1)

                                        

m=sin6x+cos6x=(sin2x+cos2x)(sin4x+cos4x-sin2xcos2x)


as a3+b3=(a+b)(a2-ab+b2)


we know that,


sin2x+cos2x=1


m=((sin2x+cos2x)2-3sin2xcos2x)


m=1-3sin2xcos2x


m=1-3/4(sin2x)2


we know 0<=sin2A<=1,


so -1<=-sin22x<=0 multiply with -1,


-3/4<=-3/4(sin22x)<=0 multiply with 3/4,


1/4<=1-3/4(sin22x)<=1 add 1,


1/4<=m<=1


therefore "m" belongs to[1,1/4].

2 years ago

Post Your Answer

More Questions On Trigonometry

Ask Experts

Have any Question? Ask Experts
Post Question
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details
if secA+tanA=x,then what are the values of sinA,secA,tanA
 
 
secA+tanA=x.........(1.) and we know that sec 2 A-tan 2 A=1 ON DIVIDING with given result with eqn.1secX-tanX=1/x(2.) now solve eqn(1.)and(2.) by simple linear equation method to obtain...
 
siddharth gupta one month ago
Whats the cosine rule.
 
 
Hello Student, Thanks & Regards Arun Kumar Btech, IIT Delhi Askiitians Faculty
  img
Arun Kumar one month ago
COS(A-B)/COS(A+B)+COS(C+D)/COS(C-D)=0 THEN SHOW THAT tan A tan B tan C tan D=-1
 
 
STEP1:COS(A-B)/COS(A+B)=(COS(A)COS(B)+SIN(A)SIN(B))/(COS(A)COS(B)-SIN(A)SIN(B)) SIMILARLY OPEN THE EXPRESSION FOR THE OTHER TERM. STEP2:DIVIDE NUMERATOR AND DENOMINATOR IN BOTH CASES BY...
 
siddharth gupta one month ago
A VECTOR OF MAGNITUDE ‘a’ IS ROTATED THROUGH ANGLE THETA.WHAT IS THE MAGNITUDE OF THE CHANGE IN THE VECTOR?
 
 
Dear Bharat As the vector has been rotated by angle theta, Let the final vector be called as v f , while the initial being v i The magnitude of change in vector is | v f – v i | = ( v i 2 +...
 
shashi K Sharma 4 days ago
 
THANKS FOR ALL THE ANSWERED PROBLEMS.
 
bharat makkar 4 days ago
integrate the following integral ∫ 1/cos^2(x)(1 – tan x)^2 dx
 
 
Ans: Hello Student, Please find answer to your question below
  img
Jitender Singh 9 days ago
All the points lying inside the triangle formed by the points (1,3), (5,6) amd (-1,2) satisfy: a)3x+2y_ 0 b) 2x+y+1_0 c) -2x+11_0 d)2x+3y-12_0
 
 
I think there is some problem with your question since the points lying inside a triangle constitute an enclosed area. So these pointscannot lie on a st. line.
 
Y RAJYALAKSHMI one month ago
View all Questions »