IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

the.sniper

The minimum value of $(1+\frac{1}{sin^n\alpha})(1+\frac{1}{cos^n\alpha})$ is ?

aditya_arora04

Hi,

I am not too sure but acc. to me, the answer's like this.

sinx and cosx can have values only,

So, max value of func. having sinx and cosx, should be at x = pie/4

Am i correct ????

EFFORT

when n is odd min. value is 0

when n is even min. value is 2

raulrag009

{iam takin alpha as @}

My try

Reduce the given xpression , u'll get somethin like this

Let S= [1/sinⁿ@cosⁿ@ + 1/cosⁿ@ + 1/sinⁿ@ + 1 ]

Let (1/sinⁿ@cosⁿ@),(1/cosⁿ@),(1/sinⁿ@),1 be terms of a prog.

AM>=GM

S>=4[1/sin@cos@]^n/2

S>=4.2^n/2/(sin2@)^n/2

max value of sin2@=1

S>=4.root(2)

the.sniper

LAMPARD

On opening the brackets,we get 1+(secx)^n +(cosecx)^n+(secx*cosecx)^n...dis exp. can be written as [1-(secx*cosecx)^n/2]^2 +(secx)^n +(cosecx)^n+2*(secx*cosecx)^n/2....dis exp. is minimum wen [1-(secx*cosecx)^n/2]^2 is minimum...(it cannot be 0)...dat happens wen x=pie/4...so min. value cums out 2 be [1-(2)^n/2]^2+[1+(2)^n/2]^2...=2+2^(n+1)...i think dis maybe de ans...

elastiboysai

Hi sniper,

here u go

hsbhatt

Assuming sin

and cos

are +ve , it is a straightforward application of Cauchy-Schwarz inequality:

(a²+b²) (c²+d²)

(ac+bd)²

If a = 1, b = 1/sin^n/2

, c = 1, d = 1/cos^n/2

then (1+1/sinⁿ

) (1+1/cosⁿ

)

(1+1/sin^n/2

cos^n/2

)² = (1+2^n/2/sin^n/22

)²

(1+2^n/2)²

Equality occurs when

= n

+

/4

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