cosx.cos2x.cos4x.cos8x.cos16x

509 Points
13 years ago

Y = cosxcos2xcos4xcos8xcos16x

taking log both sides

logY = log(cosx.cos2x.cos4x.cos8x.cos16x)

logab = loga + logb        (property)

logY =[ logcosx + logcos2x + logcos4x + logcos8x + logcos16x ]           ...................1

differentiating wrt x

differentiatn of log(f(x) = f1x/ f(x)

(1/Y) . (y1)  =

[ -sinx/cosx + -sin2x/2cos2x - sin4x/4cos4x - sin8x/8cos8x                                                 -16sin16x/16cos16x]

y1 = -Y[tanx + tan2x/2 + tan4x/4 + tan8x/8 + tan16x/16] .....................2

now put the value of Y from 1 in eq 2 u will get the desired result...

Akarsh Gaurav
24 Points
6 years ago
y=f(x)=cosx.cos2x.cos4x.cos8x.cos16x
=2sinx.cosx.cos2x.cos4x.cos8x.cos16x/2sinx                      [${\color{Red} \because }\displaystyle$multiply 2sinx/2sinx in RHS]
=2sin2x.cos2x.cos4x.cos8x.cos16x/2.2sinx                       [{sin2x=2sinx.cosx} & multiply 2/2 in                                                                                                           RHS]
SIMILARLY,
=2sin4x.cos4x.cos8x.cos16x/2.4sinx
=2sin8x.cos8x.cos16x/2.8sinx
=2sin16x.cos16x/2.16sinx
=sin32x/32sinx
${\color{Red}\therefore }\displaystyle$y=f(x)=sin32x/32sinx
differentiate both sides w.r.to x,
dy/dx=d(sin32x/32sinx)/dx
differentiate it you will getthe answer.

THANK YOU
Nitish jha
28 Points
6 years ago
The question can be written asCos2^0x.cos2^1x.cos2^2x.cos2^3x.cos2^4xHere n=4Simply put in formulaSin2^(n+1)x/2^(n+1)sinxSin2^{5}x/2^5sinxSin32x/32sinxIf u want to find exact vakue of x, differentiate y w.r.t x