MHB Can you prove \tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}?

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The discussion centers around proving the equation tan 20° + 4 sin 20° = √3. The proof involves using trigonometric identities and simplifications, ultimately showing that tan 20° can be expressed in terms of sin 20°. The calculations demonstrate that tan 60° equals √3, confirming the original equation. Participants express appreciation for each other's methods in solving the problem. The conversation highlights the collaborative nature of mathematical problem-solving.
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Prove that $$\tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}$$.
 
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anemone said:
Prove that $$\tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}$$.

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)
 
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)

Thanks for participating, kaliprasad and I like your method in cracking this problem so much!
 
anemone said:
Thanks for participating, kaliprasad and I like your method in cracking this problem so much!

Thanks anemone. As long as you ask questions whether I answer or not I shall not get Alzheimer's disease.
 

Thread 'Area of Overlapping Squares'

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