Question

show that sin 12*sin 48*sin 54=1/8

Answer 1

 Let  x = sin54 sin48 sin12  
           = (1/2) x sin 54 x ( 2 sin48 sin 12 )
We know, 2 sin A sin B =cos(A − B) − cos(A + B)
           = 1/2 x sin54 x (cos36-cos60) 
We know, sin( 90 - x ) = cos x and cos 60 = 1/2
           =1/2 x cos36 x (cos36-1/2) 
           =1/2 x (cos36)² - 1/4 x cos36

We know, cos² x = 1 + cos 2x  

           =1/4 x (1+cos72) - 1/4 x cos36 
           =1/4 + (1/4)(cos72-cos36)

We know, cos A − cos B = - 2 sin (A + B/ 2)  sin (A − B/ 2) 
 
           =1/4-1/2 x sin54 x sin18 
      We know, sin 90-x  = cos x
           =1/4-1/2 x cos36 x sin18 

We multiply both sides with 8cos18 

8cos18x = 8cos18 x (1/4-1/2 x cos36 x sin18) 
8cos18x = 2cos18 - 4cos36 x sin18 x cos18 

We know, sin 2 x =2 sin x cos x
 
8cos18x = 2cos18 - 2cos36 x sin36 
8cos18x= 2cos18-sin72 
8cos18x= 2cos18-cos18 
8cos18x = cos18

Thus, 
 x=cos18/(8cos18)
   =1/8

Answer 2

Answer:

Step-by-step explanation: