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dvrravi

sin12sin48sin54

good luck...

ankurgupta91

sin 12 sin48 sin54
= (2sin12sin48 .sin54)/2
= [(cos 36 - cos 60 ) . sin(90-36)]/2 [2sinAsinB= cos(A-B) -Cos(A+B)]
as cos 36 = (sqroot of 5 +1) /4
=[(sq root of 5 +1 )/4 -1/2 ] * [(sq root of 5 +1 )/4]/2
by solving it we get
sin 12 sin 48 sin54 =1/8
thats the answer

rajat

the answer should be 1/8

here's my solution

z = sin 12 sin48 sin 54

can be written as ...... z = sin (30-18)sin(30+18)sin(90-36)

now ,for proceeding ,u should know the value of sin 18 which is common

multiply and divide z by 2

we get 2z = [ - cos60 + cos36]cos[36]

=> 2z = [2cos36 - 1]cos36

=> 2z = 2cos^2(36) - cos36

=> 2z = 2cos^(36) - 1 + 1 - cos36

=> 2z = cos(72) + 2sin^2(18)

now cos(72) = sin(90 - 72)

=> 2z = sin18 + 2 sin^2(18)

now put value of sin18 in it u 'll get z = 1/8

i hope someone comes up with a better soln.

do rate me !!

iitkgp_bipin

Rajat is correct.

Hey Ankur....the product of three quantities less than 1 cannot be 1.
Just check your solution.

avinash.sharma

Answer 1/8 is correct :

More compact solution using only cos 36 = (

5+1)/4

sin 12 sin48 sin 54

= 1/2 (2 sin48 sin12) sin(90-36)

= 1/2( cos36 - cos60) cos 36

=1/2 ((

5+1)/4 - 1/2) . (

5+1)/4

= 1/2 {(

5-1)/4} {(

5+1)/4}

=1/2 {(5-1)/16} = 1/8

phoenix_amyth

jus try 2 make pairs bymulti n divide by 2 ...n put the value of cos36ok >>>do remember to bookmark me..bye

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