Thread: Finding x with arcsin and arccos in expression

1. 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 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. Thank you.

2. 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).

3. Re: Finding x with arcsin and arccos in expression

Hello, benny92000!

$\displaystyle x \ge 0\:\text{ and }\,\arcsin x \:=\: \arccos 2x.\;\;\text{ Solve for }x.$

Take the sine of both sides: .$\displaystyle \sin\big(\arcsin x\big) \:=\:\sin\big(\arccos 2x\big)$

. . and we have: .$\displaystyle x \;=\;\sin\big(\arccos 2x\big)$ .[1]

Let $\displaystyle \theta = \arccos2x \quad\Rightarrow\quad \cos\theta = 2x$

We have: .$\displaystyle \cos\theta \:=\:\frac{2x}{1} \:=\:\frac{adj}{hyp}$

$\displaystyle \theta$ is in a right triangle with: .$\displaystyle adj = 2x,\;hyp = 1$
Pythagorus says: .$\displaystyle opp \,=\,\sqrt{1-4x^2}$

Then: .$\displaystyle \sin\theta \:=\:\frac{opp}{hyp} \:=\:\sqrt{1-4x^2} \quad\Rightarrow\quad \theta \:=\:\arcsin\sqrt{1-4x^2}$

Hence, [1] becomes: .$\displaystyle x \;=\;\sin\big(\arcsin\sqrt{1-4x^2}\big)$

n . . . . . . . . . . . . . . . $\displaystyle x \;=\;\sqrt{1-4x^2}$

Square both sides: . . $\displaystyle x^2 \:=\:1-4x^2$

. . . . . . . . . . . . . . . $\displaystyle 5x^2 \:=\:1$

n . . . . . . . . . . . . . . $\displaystyle x^2 \:=\:\frac{1}{5}$

. . . . . . . . . . . . . . . . $\displaystyle x \:=\:\frac{1}{\sqrt{5}}$