Solve tan(30)=(sin(theta))/(1+cos(theta)) Without a Graph

[trajan22]

tan(30)=(sin(theta))/(1+cos(theta))

The only way I can solve this is by using the graph on the calculator. There must be a way to solve it by hand though but I can't find it. Maybe I am just not thinking straight but its really getting to me.

 

[Werg22]

What about

\cot (30) = \frac {1 + \cos (\theta)}{\sqrt {1 - \cos^{2} \theta}}

?

 

[Hurkyl]

Isn't that original equation exactly in the form of one of the trig identities?

 

[Werg22]

I tried to find a trigonometric identity that reduces the expression to one unknown, but couldn't.

 

[trajan22]

I think I found it. It came to me when I went to get the mail. :)
basically I have this
tan(30)=\frac {\sqrt {1 - \cos^{2} \theta}}{1+\cos(\theta)}
From here I just put it in a quadratic form and solved.

 

[Hurkyl]

The original expression is essentially one of the half-angle formulas for tangent.

 

[trajan22]

Almost, but not quite. It looks pretty tough to put it in that particular form. The way I mentioned works so I guess Ill just stick with that, especially since remembering all those identities is a pain.

Thanks for the suggestions though.

 

[theperthvan]

but not knowing them is obviously more painful.

 

[Hurkyl]

Almost, but not quite. It looks pretty tough to put it in that particular form.

It's tough to put
\tan 30^\circ = \frac{\sin \theta}{1 + \cos \theta}

into the form

\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}

?

 

[Hurkyl]

I think I found it. It came to me when I went to get the mail. :)
basically I have this
tan(30)=\frac {\sqrt {1 - \cos^{2} \theta}}{1+\cos(\theta)}
From here I just put it in a quadratic form and solved.

By the way, that expression isn't right -- you don't know that the numberator is the positive root. You also have to consider

tan(30)=-\frac {\sqrt {1 - \cos^{2} \theta}}{1+\cos(\theta)}

 

[trajan22]

It's tough to put
\tan 30^\circ = \frac{\sin \theta}{1 + \cos \theta}

into the form

\tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}

?

Unless I am missing something I thought the half angle formula was <br /> <br /> \tan \frac{\theta}{2} = \frac{\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}}

Thats not the same as \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} right? Or am I making a mistake?

 

[Werg22]

And just what multiplying \frac{\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}} by

\frac {\sqrt{ 1 + \cos \theta}}{\sqrt{ 1 + \cos \theta}}

gives?

 

[VietDao29]

And just what multiplying \frac{\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}} by

\frac {\sqrt{ 1 + \cos \theta}}{\sqrt{ 1 + \cos \theta}}

gives?

Well, it gives:
... = \frac{\sqrt{1 - \cos ^ 2 \theta}}{1 + \cos \theta} = \frac{\textcolor{red} {|} \sin \theta \textcolor{red} {|}}{1 + \cos \theta} (Notice that it's never negative, since there's an absolute value in it)

What you should use is Half-Angle and Power Reduction Identities:

\sin (2 \theta) = 2 \sin ( \theta ) \cos ( \theta )

and: \cos ^ 2 \theta = \frac{1 + \cos (2 \theta)}{2}

 

[Werg22]

?

One solution is found by taking the positive root and the other by taking the negative, as such

\tan \frac{\theta}{2} = \frac{-\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}}

 

[VietDao29]

?

One solution is found by taking the positive root and the other by taking the negative, as such

\tan \frac{\theta}{2} = \frac{-\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}}

What Hurkyl, and I are (?, is it is, am, or are should be used here?) trying to say is that, the expression:
\tan \left( \frac{\theta}{2} \right) = \frac{\sqrt{1 - \cos \theta}}{\sqrt{1 + \cos \theta}} is, indeed, incorrect.

It'll be correct if you take into account its negative part as well, i.e using the (+/-) sign as you did.

 

[Werg22]

What Hurkyl, and I are (?, is it is, am, or are should be used here?) trying to say is that, the expression:
\tan \left( \frac{\theta}{2} \right) = \frac{\sqrt{1 - \cos \theta}}{\sqrt{1 + \cos \theta}} is, indeed, incorrect.

It'll be correct if you take into account its negative part as well, i.e using the (+/-) sign as you did.

Technicalities, you say tomato, I say tomato.

 

[pradeep reddy]

tan(30)=(sin(theta))/(1+cos(theta))

The only way I can solve this is by using the graph on the calculator. There must be a way to solve it by hand though but I can't find it. Maybe I am just not thinking straight but its really getting to me.

tan30=sin(@)/(1+cos@)
=2sin(@\2)cos(@\2)/2cossquare(@/2)
=tan(@/2)
implies 30=@/2
@=60

here @=theta

 

[Hurkyl]

@=60

Of course, that's merely one solution.

(p.s. use [ code ] ... [ /code ] tags for preformatted text)