# Trig function of arc trig functions and the reverse

1. Sep 1, 2011

### Philosophaie

I know the sin(arccos(x)) = (1-x^2)^0.5

I was wondering what some of the others are:

cos(arcsin(X))
tan(arcsin(X))
tan(arccos(x))
sin(arctan(x))
cos(arctan(x))

also the reverse:

arcsin(cos(x))
arcsin(tan(X))
arccos(Sin(X))
arccos(tan(X))
arctan(sin(X))
arctan(cos(X))

2. Sep 1, 2011

### micromass

Staff Emeritus
Let me do two:

$$\cos(arcsin(x))=\sqrt{1-\sin(arcsin(x))}=\sqrt{1-x^2}$$

and

$$\tan(arccos(x))=\sqrt{\frac{1}{\cos^2(arccos(x))}-1}=\sqrt{\frac{1}{x^2}-1}$$

I'll let you find out the other ones...

3. Sep 1, 2011

### gsal

are these always possible? I mean take

arccos(tan(x))

for example, the cosine of an angle is always between 0 and 1, and so, the argument to the arccos function should be a number between 0 and 1...but the tangent of an angle can get pretty large...so, I think these is no solution here...same for others.

4. Sep 1, 2011

### LCKurtz

Not if arccos(x) is in the second quadrant.

5. Sep 2, 2011

### HallsofIvy

Staff Emeritus
As long as your "angles" are in the first quadrant (so you don't have multi-value problems), you can get all of those formulas by constructing an appropriate right triangle.

For example, to get sin(arctan(x)), imagine a right triangle with "opposite side" x and "near side" 1 (so that the tangent of the angle opposite side "x" is x/1= x and the angle is arctan(x)). By the Pythagorean theorem, it will have "hypotenuse" $\sqrt{x^2+ 1}$. Sine is "opposite side over hypotenuse" so $sin(arctan(x))= \frac{x}{\sqrt{x^2+ 1}}$.