Question

Sin π/18 Sin5π/6 sin 5π/18 Sin 7π/18 =1/16​

Answer 1

||✪✪ CORRECT QUESTION ✪✪||

• Prove that Sin π/18 × Sin3π/18 × sin 5π/18 × Sin7π/18 = 1/16

|| ✰✰ ANSWER ✰✰ ||

We know that ,

π/18 = 10° , 3π/18 = 30° , 5π/18 = 50° and 7π/18 = 70°..

So,

in LHS we have ,

➪ sin10° × sin30° × sin50° × sin70°

We know that sin30° = 1/2 .

______________________

So, Lets Try to Solve Rest part :-

➺ sin 10 × sin 50 × sin 70

➺ ( sin 10 × sin 50 ) × sin 70

using [ sinA × sinB = (-1/2) { cos(A+B) - cos(A-B) } ] we get

➺ (-1/2) [ cos (10+50) - cos (50-10) ] × sin 70

➺ (-1/2) [ cos 60 - cos 40 ] × sin 70

Putting cos60° = 1/2 Now,

➺ (-1/2) [ 1/2 - cos 40 ] × sin 70

➺ (-1/4) [ 1 - 2cos 40 ] × sin 70

➺ (-1/4) [ sin 70 - 2 cos 40 × sin 70 ]

Using cosA = Sin(90-A) now,

➺ (-1/4) [ sin 70 - 2 sin( 90 - 40 ) × sin 70 ]

➺ (-1/4) [ sin 70 - 2 sin 50 × sin 70 ]

using 2SinA×SinB = (-1){ cos(A+B) - cos(A-B) }

➺ (-1/4) [ sin 70 + ( cos( 50 + 70 ) - cos( 70 - 50) ) ]

➺ (-1/4) [ sin 70 + ( cos 120 - cos 20 ) ]

Again , using cosA = Sin(90-A) now

➺ (-1/4) [ sin 70 + ( cos 120 - sin ( 90 - 20 ) ) ]

➺ (-1/4) [ sin 70 + ( cos 120 - sin 70 ) ]

➺ (-1/4) [ sin 70 + cos 120 - sin 70 ]

➺ (-1/4) [ cos 120 ]

Putting cos120° = (-1/2) now,

➺ (-1/4) × ( -1/2 )

➺ 1/8

_________________

Now, In LHS, we have ,

☛ sin30° × [ sin10° × sin50° × sin70° ]

☛ 1/2 × 1/8

☛ 1/16 = ❦❦ RHS ❦❦

✪✪ Hence Proved ✪✪

Answer 2

QUESTION :- Prove that Sin π/18 × Sin3π/18 × sin 5π/18 × Sin7π/18 = 1/16

ANSWER :-

π/18 = 10° , 3π/18 = 30° , 5π/18 = 50° and 7π/18 = 70°..

→ sin30° * sin10° * sin50° * sin70°

as sin30° = 1/2 , solving rest ,

= sin10° * sin50° * sin70°

= (-1/2) (cos60 - cos40)sin70

= (-1/2) (1/2 - cos40)sin70

= (-1/4) (1 - 2cos40)sin70

= (-1/4)(sin70-2cos40sin70)

= (-1/4)(sin70-2sin50sin70)

= (-1/4)(sin70+(cos120-cos20)

= (-1/4)(sin70-cos20-1/2)

= (-1/4)(sin70-sin70-1/2)

= (-1/4)(-1/2)

= 1/8 .

So,

sin30° * sin10° * sin50° * sin70°

→ 1/2 * 1/8

→ 1/16 (RHS) Proved ...