Finding x with arcsin and arccos in expression

If x is greater than or equal to 0 and arcsinx = arccos x (2x), then x=?

The answer is .447.

The book says to take the sine of both sides, then apply pythagorean identity. I know the pythagorean identity, but I don't understand how it applies here. (Headbang) Thank you.

Re: Finding x with arcsin and arccos in expression

You draw a Right Triangle. Designate one acure angle as the angle of interest, call is S.

When contemplating sin(arccos(x)), notice that arccos(x) is an ANGLE. Let's call it S

If the cos(S) = x, then you can lable the leg adjacent to S, just "x" and the hypotenuse "1".

The the Pythagorean identity sugests the opposite side is sqrt(1^2 - x^2)

Simply write down sin(S).

Re: Finding x with arcsin and arccos in expression

Hello, benny92000!

Take the sine of both sides: .

. . and we have: . .[1]

Let

We have: .

is in a right triangle with: .

Pythagorus says: .

Then: .

Hence, [1] becomes: .

n . . . . . . . . . . . . . . .

Square both sides: . .

. . . . . . . . . . . . . . .

n . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . .