Finding the Angles: A Trigonometric Problem

[omicron]

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

 

[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.

 

[omicron]

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

 

[Nylex]

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

 

[Kahsi]

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

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

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

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

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

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

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

 

[omicron]

Don't know

 

[Nylex]

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

Can you go from there?

 

[jai6638]

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

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

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

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

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

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

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

How did you get all that??

 

[Zurtex]

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

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

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

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

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

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

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

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:

$$x - 30 = 50$$

 

[whozum]

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

 

[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

$$x = 80, 260$$

 

[omicron]

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

Can you go from there?

Yup. Thanks!

 

[Kahsi]

It's not that complex:

Hence my smilies

 

[Zurtex]

Hence my smilies

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:

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