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I have a question that asks me to find an algebraic expression for sin(arccos(x)). From the lone example in the book I seen they're doing some multistep thing with the identities, but I'm just not even sure where to start here. It's supposed to be $\sqrt{1-x^2}$

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What is $\arccos(x)$? it is the angle given by the ratio of sides of length $x$ and $1$. That is, on a triangle with adjacent side length $x$ and hypotenuse length $1$ we will find the angle $\arccos(x)$.

(There is a picture for this which I hope you can get from what I describe).

Now what is $\sin(y)$? It is the ratio of the opposite side to the hypotenuse of a triangle with angle $y$ in the respective part. So if we use $y=\arccos(x)$, we have that $\sin(y)$ is the ratio of the sides with degree given by a triangle whose adjacent and hypotenuse sides are length $x$ and $1$ respectively.

Since we have the adjacent and hypotenuse lengths, we can calculate the opposite sides length by the Pythagorean theorem. This means, if we use $z$ for the opposite side, that $z^2+x^2=1^2=1$. Solving for $z$ gives $z=\sqrt{1-x^2}$.

Then $\sin(y)$ is the ratio of the opposite size, $z=\sqrt{1-x^2}$ and the hypotenuse, $1$. We may now say $\sin(arcos(x))=\sqrt{1-x^2}$.

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Set $\alpha=\arccos x$. Since, by definition, $\alpha\in[0,\pi]$, you know that $\sin\alpha\ge0$, so $$ \sin\arccos x=\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-(\cos\arccos x)^2}=\sqrt{1-x^2} $$

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Let $\arccos(x)=\theta$

$x=\cos(\theta)$

You require $y=\sin(\arccos(x))=\sin(\theta)$

Identity: $\cos^2(\theta)+\sin^2(\theta)=1$

$x^2+y^2=1$

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Draw a right triangle with one of the angles being $cos^{-1} x$, then fill out the lengths of the sides of the triangle using the definition of cosine for two of the sides, and Pythagoras's theorem for the third side. Then, use the definition of sine to find $sin(cos^{-1} x)$. For questions like these, always try drawing a triangle first.

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$$y=\sin\arccos x$$ $$x=\cos\arccos x$$

Therefore $x$ and $y$ are the sin and cos of the same angle and

$$x^2+y^2=1$$

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