- #1

- 461

- 0

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))

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Philosophaie
- Start date

- #1

- 461

- 0

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

- 22,129

- 3,297

[tex]\cos(arcsin(x))=\sqrt{1-\sin(arcsin(x))}=\sqrt{1-x^2}[/tex]

and

[tex]\tan(arccos(x))=\sqrt{\frac{1}{\cos^2(arccos(x))}-1}=\sqrt{\frac{1}{x^2}-1}[/tex]

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

- #3

- 1,065

- 54

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

- 9,559

- 770

Let me do two:

[tex]\tan(arccos(x))=\sqrt{\frac{1}{\cos^2(arccos(x))}-1}=\sqrt{\frac{1}{x^2}-1}[/tex]

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

- #5

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 965

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" [itex]\sqrt{x^2+ 1}[/itex]. Sine is "opposite side over hypotenuse" so [itex]sin(arctan(x))= \frac{x}{\sqrt{x^2+ 1}}[/itex].

Share: