# Is this solvable

1. Jun 8, 2007

### 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 cant find it. Maybe Im just not thinking straight but its really getting to me.

2. Jun 8, 2007

### Werg22

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

?

3. Jun 8, 2007

### Hurkyl

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

4. Jun 8, 2007

### Werg22

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

5. Jun 8, 2007

### 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.

6. Jun 8, 2007

### Hurkyl

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

7. Jun 8, 2007

### 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.

8. Jun 8, 2007

### theperthvan

but not knowing them is obviously more painful.

9. Jun 9, 2007

### Hurkyl

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

into the form

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

?

Last edited: Jun 9, 2007
10. Jun 9, 2007

### Hurkyl

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

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

11. Jun 9, 2007

### trajan22

Unless Im missing something I thought the half angle formula was $$\tan \frac{\theta}{2} = \frac{\sqrt {1-\cos \theta}}{\sqrt{ 1 + \cos \theta}}$$

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

Last edited: Jun 9, 2007
12. Jun 9, 2007

### 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?

13. Jun 10, 2007

### VietDao29

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

Last edited: Jun 10, 2007
14. Jun 10, 2007

### Werg22

???

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

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

15. Jun 10, 2007

### VietDao29

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.

Last edited: Jun 10, 2007
16. Jun 12, 2007

### Werg22

Technicalities, you say tomato, I say tomato.

17. Oct 7, 2007

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

here @=theta

18. Oct 7, 2007

### Hurkyl

Staff Emeritus
Of course, that's merely one solution.

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