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

[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.

 

[CellCoree]

I cannot "prove" mathematical equations in the same way that I can prove scientific theories through experimentation and observation. However, I can provide evidence and reasoning to support the validity of this equation.

To begin, we can use the trigonometric identity \tan \theta = \frac{\sin \theta}{\cos \theta} to rewrite the left side of the equation as \frac{\sin 20^{\circ}}{\cos 20^{\circ}} + 4\sin 20^{\circ}. We can then factor out a common term of \sin 20^{\circ} to get \sin 20^{\circ} \left(\frac{1}{\cos 20^{\circ}} + 4\right).

Next, we can use the Pythagorean identity \cos^2 \theta + \sin^2 \theta = 1 to solve for \cos 20^{\circ}. We get \cos 20^{\circ} = \sqrt{1 - \sin^2 20^{\circ}}. Substituting this into our equation, we get \sin 20^{\circ} \left(\frac{1}{\sqrt{1 - \sin^2 20^{\circ}}} + 4\right).

We can then use the trigonometric identity \sin^2 \theta + \cos^2 \theta = 1 to rewrite the expression inside the parentheses as \frac{1 + \sin^2 20^{\circ}}{\sqrt{1 - \sin^2 20^{\circ}}}.

Using the Pythagorean identity again, we can simplify this to \frac{1 + \sin^2 20^{\circ}}{\cos 20^{\circ}}. Substituting this back into our original equation, we get \sin 20^{\circ} \cdot \frac{1 + \sin^2 20^{\circ}}{\cos 20^{\circ}}.

We can then use the trigonometric identity \sin^2 \theta = 1 - \cos^2 \theta to rewrite this as \frac{\sin 20^{\circ} \cdot \cos^2 20^{\circ}}{\cos 20^{\circ}}.

Finally, we can cancel out the common term of \cos 20^{\circ} to get \sin 20^{\