# [/tex]

1
by karthik4297
yes pi into cosx is inside in sin function
are you sure denominator has x power 4? is it x square? why do u write root of 1/4? canot you write simply 1/2 ? is there something missing?
yes denominator has x^4
ans is pi^2
check it now. done

2014-08-01T00:45:21+05:30

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Sin² [ πcos² x ] = sin² [ π(1+cos 2x)/2 ]  = sin² [ π/2 + π/2 cos2x]
= cos² [π/2 cos 2x ]  = 1/2 * [ 1 + cos ( π cos 2x ) ]
t = π cos 2x            as x -> 0,  t -> π
x =  1/2 *  Cos^-1  ( t/π )
denominator 1/2 x^4

===========================================
Try in some other way:  Let us use the Taylor series of expansion for cos x.

cos x = 1 - x² /2 + x^4 /4! - x^6 /6!  + ..  = 1 - x² y      where y = 1/2 - x²/4! + x^4/6! - ...
y -> 1/2 as x->0
So  cos² x = (1 + x^4  y² - 2 x² y)

Sin² [ πcos² x ] = sin² [ π + πx^4 y² - 2 π x² y ]                        sin π+t = - sin t
= sin ² [ πx^4 y²  - 2 π x² y ]
Now problem is    lim as x-> 0

sin² [πx^4 y²  - 2 π x² y ]
=    ------------------------------
x^4
sin² [πx^4 y² - 2 π x² y ]      [πx^4 y² - 2 π x² y ]²
=        ----------------------------------  *  ----------------------                  multiply & divide by same.
[πx^4 y² - 2 π x² y ]²                    x^4

=     1² *   (πx²y)² [x² y² - 2]²  / x^4   =    π² y² [ x² y² - 2]²
= as x-> 0  this is          =   π² y² (-2)² =     π² (1/2)² (4) =  π²
answer seems to be  π²

it means ?
answer is pi square