Question

Show that sin12º.sin48º.sin54° = 1/8

Answer 1

Step-by-step explanation:

 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