# Tan x = cos x

1. Sep 13, 2010

### Helios

If tan x = cos x, then what is x ? The answer includes the golden ratio !

2. Sep 13, 2010

### Char. Limit

3. Sep 13, 2010

So let s = sin x, c = cos x; then s/c = c, so s = c2 = 1 - s2. Solve for s, then take arcsin. Easy.

4. Sep 13, 2010

### Char. Limit

To be exact...

$$Answer = sin^{-1}\left(\phi\right)$$

5. Sep 13, 2010

You forgot the other solution: $$\sin^{-1}\bigl[-\tfrac12(1+\sqrt5)\bigr]$$. (Not a real solution, though.)

6. Sep 13, 2010

### Char. Limit

I tend to use the real numbers.

Complex numbers are not often the solutions to questions found in the general math section.

EDIT: I just realized that $sin^{-1}(\phi)$ isn't real either... wow, so there are actually no real solutions.

7. Sep 13, 2010

### JJacquelin

Tan(x) = Cos(x)
Real solution are :
x = ArcSin((Sqrt(5)-1)/2) + 2*k*Pi
and
x = Pi - ArcSin((Sqrt(5)-1)/2) + 2*k*Pi

k = any négative, nul or positive integer.

8. Sep 13, 2010

### Char. Limit

So, in other words, any integer?

9. Sep 13, 2010

### JJacquelin

Yes, indeed !

10. Sep 13, 2010

### Mentallic

So cos(x) doesn't cross tan(x)?

11. Sep 13, 2010

### Char. Limit

Mistake number two...

The real solution is in fact:

$$sin^{-1}\left(\frac{-1}{\phi}\right)$$

I think, assuming that

$$\frac{-1}{\phi} = \frac{1-\sqrt{5}}{2}$$

Which I seem to remember a similar property about phi...

12. Sep 13, 2010

### Helios

note that

tan( x ) = cos( x ) = 1 / sqr( phi )

so the solution can be made in terms of arctan or arccos.

13. Sep 15, 2010

### willem2

Since tan(x) increases monotonically from 0 to infinity in the domain (0, pi/1) and
cos (x) decreases from 1 to 0 in the same domain, there should be a solution x between 0 and pi/2 and 0< sin(x) < 1. There's a solution in the second quadrant as well. cos(x) and tan(x) have different signs in the 3rd and fourth quadrant, so a solution where sin(x) < 0 is not possible

Once you get all the signs right in the quadratic formula, you'll get

$$sin(x) = \frac {1} {\phi} = \phi - 1$$

from one of the solutions. The other solution has sin(x) >1 so that isn't a valid solution.

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