Prove that tan(20)+4sin(20)=√3?

Identities used:
sin(2A) = 2 sin(A) cos(A)
sin(A - B) = sin(A) cos(B) - cos(A) sin(B)

sin(60°) = √3/2  and  cos(60°) = 1/2

L.H.S.
= tan(20°) + 4 sin(20°)
= [sin(20°)/cos(20°)] + [4 sin(20°) cos(20°) / cos(20°)]
= [1/cos(20°)] [sin(20°) + 4 sin(20°) cos(20°)]
= [1/cos(20°)] [sin(20°) + 2 {2 sin(20°) cos(20°)}]
= [1/cos(20°)] [sin(20°) + 2 sin(40°)]
= [1/cos(20°)] [sin(20°) + 2 sin(60° - 20°)]
= [1/cos(20°)] [sin(20°) + 2 sin(60°) cos(20°) - 2 cos(60°) sin(20°)]
= [1/cos(20°)] [sin(20°) + 2 (√3/2) cos(20°) - 2 (1/2) sin(20°)]
= [1/cos(20°)] [sin(20°) + (√3) cos(20°) - sin(20°)]
= [1/cos(20°)] [(√3) cos(20°)]
= √3
= R.H.S.

Hence, tan(20°) + 4 sin(20°) = √3

Let T = tan(20) + 4sin(20)
Let t, s, c refer to 20 degrees and let S refer to 40 degrees
T = t + 4s = s/c + 4s = (1/c)[s + 4sc] = (1/c)[s + 2S]
T = (1/c)[s + 2 sin(60 - 20)]] ..................(1)
2sin(60 - 20) = 2sin(60)cos(20) - 2cos(60) sin(20)
2sin(60 - 20) = √(3)*c – s
T = (1/c)[s +√(3)*c – s] = √(3)

micatkie posted this approach first

Recall the identity:

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = y and suppose that: b = x

sin(x + x) = sin(x).cos(x) + cos(x).sin(x)

sin(2x) = 2.sin(x).cos(x)

2.sin(x).cos(x) = sin(2x)

4.sin(x).cos(x) = 2.sin(2x) ← memorize this result as (1)


Recall the identity:

sin(a - b) = sin(a).cos(b) - cos(a).sin(b) → suppose that: a = 60 and suppose that: b = 20

sin(60 - 20) = sin(60).cos(20) - cos(60).sin(20)

sin(40) = [(√3)/2].cos(20) - (1/2).sin(20)

2.sin(40) = √3.cos(20) - sin(20) ← memorize this result as (2)


= tan(20) + 4.sin(20)

= [sin(20)/cos(20)] + 4.sin(20)

= [sin(20) + 4.sin(20).cos(20)] / cos(20) → recall (1): 4.sin(20).cos(20) = 2.sin(40)

= [sin(20) + 2.sin(40)] / cos(20) → recall (2): 2.sin(40) = √3.cos(20) - sin(20)

= [sin(20) + √3.cos(20) - sin(20)] / cos(20)

= [√3.cos(20)] / cos(20)

= √3