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Answers
Step-by-step explanation:
To Prove :
32√3.sin(π/48).cos(π/48).cos(π/24).cos(π/12).cos(π/6) = 3
Solution :-
Taking L.H.S :-
= 32√3.sin(π/48).cos(π/48).cos(π/24).cos(π/12).cos(π/6)
= 16×2×√3.sin(π/48).cos(π/48).cos(π/24).cos(π/12).cos(π/6)
= 16√3.(2sin(π/48).cos(π/48))).cos(π/24).cos(π/12).cos(π/6)
= 16√3.(sin2(π/48)).cos(π/24).cos(π/12).cos(π/6)
= 16√3.sinπ/24.cos(π/24).cos(π/12).cos(π/6)
= 8√3.2.sinπ/24.cos(π/24).cos(π/12).cos(π/6)
= 8√3.sin2(π/24).cos(π/12).cos(π/6)
= 8√3.sin(π/12).cos(π/12).cos(π/6)
= 4√3.2.sin(π/12).cos(π/12).cos(π/6)
= 4√3.sin2(π/12).cos(π/6)
= 4√3.sin(π/6).cos(π/6)
= 2√3.2.sin(π/6).cos(π/6)
= 2√3.sin2(π/6)
= 2√3.sinπ/3
= 2√3 × sin60°
= 2√3 × √3/2
= √3 × √3
= 3
= R.H.S
Hence Proved.
Formula Required :-
1) sin2A = 2sinAcosA
2) π/3 = 60°
3) sin60° = √3/2
Prove that,
Consider, LHS
can be rewritten as
Let assume that,
So, above expression can be rewritten as
We know,
So, using this identity, we get
Hence,