The equation $$\tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}$$ is proven by manipulating trigonometric identities. The proof utilizes the relationship between tangent and sine functions, specifically $$\tan 60^{\circ} - \tan 20^{\circ}$$, leading to the conclusion that $$4 \sin 20^{\circ} + \tan 20^{\circ} = \tan 60^{\circ}$$, which equals $$\sqrt{3}$$. The method involves applying the sine difference formula and simplifying the expression to demonstrate the equality definitively.
PREREQUISITESMathematicians, students studying trigonometry, educators teaching trigonometric identities, and anyone interested in solving trigonometric equations.
anemone said:Prove that $$\tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}$$.
kaliprasad said:we have tan 60 - tan 20
=sin 60/cos 60 - sin 20/cos 20
= (sin 60 cos 20 - cos 60 sin 20)/ ( cos 60 cos 20)
= sin (60-20)/ (cos 60 cos 20)
= sin 40/ ( cos 60 cos 20)
= (2 sin 20 cos 20)/( cos 60 cos 20)
= 2 sin 20 / cos 60
= 2 sin 20/ (1/2) = 4 sin 20
hence 4 sin 20 + tan 20 = tan 60 = 3^(1/2)
anemone said:Thanks for participating, kaliprasad and I like your method in cracking this problem so much!