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Grade: 11
        

cosx.cos2x.cos4x.cos8x.cos16x

8 years ago

Answers : (3)

vikas askiitian expert
509 Points
							

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...


8 years ago
Akarsh Gaurav
24 Points
							
y=f(x)=cosx.cos2x.cos4x.cos8x.cos16x
         =2sinx.cosx.cos2x.cos4x.cos8x.cos16x/2sinx                      [{\color{Red} \because }\displaystylemultiply 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 }\displaystyley=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
2 years ago
Nitish jha
28 Points
							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
						
2 years ago
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