Question

prove that sin12 sin48 sin54=1÷8​

Answer 1

sin(12°)sin(48°)sin(54°)

= [sin(12°)sin(48°)]sin(54°)

= sin(54°) * [cos(48° - 12°) - cos(48° + 12°)]/2

= sin(54°) * [cos(36°) - cos(60°)]/2

= sin(54°) * [cos(36°) - 1/2]/2

= cos(36°) * [cos(36°)/2 - 1/4]

= cos²(36°)/2 - cos(36°)/4.

Since cos(36°) = (1 + √5)/4:

cos²(36°)/2 - cos(36°)/4

= [(1 + √5)/4]²/2 - [(1 + √5)/4]/4

= (1 + √5)²/32 - (1 + √5)/16

= (6 + 2√5)/32 - (1 + √5)/16

= (6 + 2√5)/32 - (2 + 2√5)/32

= 4/32

= 1/8.

So sin(12°)sin(48°)sin(54°) = 1/8 follows.

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Answer 2

HEY GUYS ,

THE ANSWER OF THE QUESTION IS AS FOLLOWS

Step-by-step explanation:

sin(12°)sin(48°)sin(54°)

= [sin(12°)sin(48°)]sin(54°)

= sin(54°) * [cos(48° - 12°) - cos(48° + 12°)]/2

= sin(54°) * [cos(36°) - cos(60°)]/2

= sin(54°) * [cos(36°) - 1/2]/2

= cos(36°) * [cos(36°)/2 - 1/4]

= cos²(36°)/2 - cos(36°)/4.

Since cos(36°) = (1 + √5)/4:

cos²(36°)/2 - cos(36°)/4

= [(1 + √5)/4]²/2 - [(1 + √5)/4]/4

= (1 + √5)²/32 - (1 + √5)/16

= (6 + 2√5)/32 - (1 + √5)/16

= (6 + 2√5)/32 - (2 + 2√5)/32

= 4/32

= 1/8.

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