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sin 36^o sin 72^o sin 108^o sin 144^o =...

`sin 36^o sin 72^o sin 108^o sin 144^o =`

A

`(1)/(4)`

B

`(1)/(16)`

C

`(3)/(4)`

D

`(5)/(16)`

Text Solution

Verified by Experts

The correct Answer is:
D
`sin36^(@)sin72^(@)sin108^(@)sin144^(@)=sin^(2)36^(@)sin^(2)72^(@)`
`=(1)/(4)[(2sin^(2)36^(@))(2sin^(2)72^(@))]`
`=(1)/(4)[(1-cos72^(@))(1-cos144^(@))]`
`=(1)/(4)[(1-sin18^(@))(1+cos36^(@))]`
`=(1)/(4)[(1-(sqrt(5)-1)/(4))(1+(sqrt(5)+1)/(4))]=(5)/(16)`
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sin 36^(º)sin 72^(º) sin108^(º) sin 144^(º) =

sin36^(@)sin72^(@)sin108^(@)sin144^(@)=

sin36^(@)sin72^(@)sin108^(@)sin144^(@)=

sin36^(@)sin72^(@)sin108^(@)sin144^(@)=(5)/(16)

Prove that: "sin"(pi)/(5)"sin"(2pi)/(5)"sin"(3pi)/(5)"sin"(4pi)/(5)=(5)/(16) "sin"36^(@) "sin"72^(@) "sin"108^(@) sin 144^(@)=(5)/(16) .

sin36^(@).sin72^(@).sin108^(@).sin144^(@)=?

Prove that: sin36^(@)s in72^(0)s in108^(0)s in144^(0)=(5)/(16)

MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-FACTORIZATION FORMULAE -EXERCISE 2
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