Trigonometric Challenge

[anemone]

Prove that \(\displaystyle \tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}\).

 

[kaliprasad]

Prove that \(\displaystyle \tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}\).

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)

 

[anemone]

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!

 

[kaliprasad]

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.