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find the value of sin18^0 and cos36^0...

find the value of `sin18^0` and `cos36^0`

A

4

B

2

C

1

D

`1//4`

Text Solution

Verified by Experts

The correct Answer is:
D
We know that `sin18^(@)=(sqrt(5)-1)/(4)`
Now , `cos36^(@)=1-2sin^(2)18^(@)`
`=1-(2)/(16)(sqrt(5)-1)^(2)=1-((3-sqrt(5)))/(4)`
`cos36^(@)=(sqrt(5)+1)/(4)`
Now , `sin18^(@)cos36^(@)=((sqrt(5))^(2)-(1)^(2))/(16)=(4)/(16)=(1)/(4)`
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