Welcome to our community

Be a part of something great, join today!

Trigonometric Challenge

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
Prove that \(\displaystyle \tan 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}\).
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
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)
 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
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

Well-known member
Mar 31, 2013
1,309
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.