Find all the angles problem

1. May 21, 2005

omicron

Find all the angles from $$0^{\circ}$$ to $$360^{\circ}$$ inclusive which satisfy the equation
$$\tan(x-30^{\circ}) - \tan 50^{\circ} = 0$$

2. May 21, 2005

Nylex

If you want help, you need to show us what you've done so far, or what your thoughts are on how to go about solving it.

3. May 21, 2005

omicron

I haven't done anything. I don't have a clue what to do.

4. May 21, 2005

Nylex

Here's a hint: write tan (x - 30) in terms of tan (50). What can you see then?

5. May 21, 2005

Kahsi

$$\tan(\alpha + \beta) = \frac{tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$$

$$\tan(x - 30) = \frac{tan(x)+\tan(-30)}{1-\tan(x)\tan(-30)}$$

$$\frac{tan(x)+\tan(-30)}{1-\tan(x)\tan(-30)}=\tan(50)$$

$$\tan(x)+\tan(-30)=\tan(50) - \tan(x)\tan(-30)\tan(50)$$

$$\tan(x) + \tan(x) \tan(-30)\tan(50)=\tan(50) - \tan(-30)$$

$$\tan(x)(1 +\tan(-30)\tan(50))=\tan(50) - \tan(-30)$$

$$\tan(x)=\frac{\tan(50) - \tan(-30)}{1 +\tan(-30)\tan(50)}$$

Last edited: May 21, 2005
6. May 21, 2005

omicron

Don't know :uhh:

7. May 21, 2005

Nylex

$$\tan(x-30^{\circ}) = \tan 50^{\circ}$$

Can you go from there?

8. May 21, 2005

jai6638

How did you get all that??

9. May 21, 2005

Zurtex

:rofl: It's not that complex:

$$\tan (x - 30) = \tan 50$$

Hence to work out an initial value just apply arctan on both sides to get:

$$x - 30 = 50$$

10. May 21, 2005

whozum

There are two solutions to the problem, that is one of them.

11. May 21, 2005

ivans_dc

since the Tan curve goes in a period of 180 degrees, you take the value that you got as one of the solutions and add or subtract 180 to/from it, and every time the result is within the rang of 0 -360, so:

you do
$$\tan (x - 30) = \tan 50$$
$$x = 80$$

then

$$80 \pm 180n = x$$

and the only other value that fits into the range is when

$$n = 1$$
$$80 + 180 = 260$$

Therefor the 2 answers are

$$x = 80, 260$$

12. May 22, 2005

omicron

Yup. Thanks!

13. May 22, 2005

Kahsi

Hence my smilies

14. May 22, 2005

Zurtex

Which if you expand is:

$$\frac{\tan(50) - \tan(-30)}{1 +\tan(-30)\tan(50)}= \frac{8 \cos^7 (10) \sin (10) - 56 \cos^5 (10) \sin^3 (10) + 56 \cos^3 (10) \sin^5 (10) - 8 \cos (10) \sin^7 (10)}{\cos^8(10) - 28 \cos^6(10) \sin^2(10) + 70 \cos^4(10) \sin^4(10) - 28 \cos^2 (10) \sin^6(10) + \sin^8 (10)}$$

And it just so happens that nicely simplifies down to:

$$\frac{8 \cos^7 (10) \sin (10) - 56 \cos^5 (10) \sin^3 (10) + 56 \cos^3 (10) \sin^5 (10) - 8 \cos (10) \sin^7 (10)}{\cos^8(10) - 28 \cos^6(10) \sin^2(10) + 70 \cos^4(10) \sin^4(10) - 28 \cos^2 (10) \sin^6(10) + \sin^8 (10)} = \tan (80)$$