## Find tan(arccos[x+1]) w/o trigonometric functions or inve

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
king&i
Posts: 23
Joined: Sat Mar 07, 2009 5:13 pm

### Find tan(arccos[x+1]) w/o trigonometric functions or inve

Find an expression for tan(arccos[x+1]) that does not involve trigonometric functions or their inverses

If anyone can help, I would really appreciate it.

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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### Re: Find tan(arccos[x+1]) w/o trigonometric functions or inve

Draw a right triangle. The particular shape doesn't matter.
```right triangle:
/|
/   |
/      |
/         |
/            |
/               |
/                  |
----------------------```
By definition, the arccosine returns an angle value. So arccos(x + 1) = @ for some angle @. (I'm using "@" to stand for "theta".) You need to find the tangent of that angle. So let's label the triangle with the angle:
```right triangle:
/|
/   |
/      |
/         |
/            |
/               |
/@                 |
----------------------```
We also know, by definition, that "arccos(x + 1) = @" means that "cos(@) = x + 1". In this case, it will be more useful to express this as "(x + 1)/1", because that gives us our "adjacent" (namely, x + 1) and our "hypotenuse" (namely, 1). So now we have:
```right triangle:
/|
/   |
1    /      |
/         |
/            |
/               |
/@                 |
----------------------
x + 1```
Use the Pythagorean Theorem to find the value for "opposite", and then read off the value of the tangent.