Expand sin(arcsin(x)-arccos(x))

The functions sine and arcsine are inverses.

The functions cosine and arccosine are inverses.

Multiply ${\displaystyle x}$ by ${\displaystyle x}$.

Draw a triangle in the plane with vertices ${\displaystyle (\sqrt{1^2-x^2},x)}$, ${\displaystyle (\sqrt{1^2-x^2},0)}$, and the origin. Then ${\displaystyle \arcsin \left(x\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (\sqrt{1^2-x^2},x)}$. Therefore, ${\displaystyle \cos \left(\arcsin \left(x\right)\right)}$ is ${\displaystyle \frac{\sqrt{1^2-x^2}}{1}}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=1}$ and ${\displaystyle b=x}$.

Divide ${\displaystyle \sqrt{(1+x)(1-x)}}$ by ${\displaystyle 1}$.

Draw a triangle in the plane with vertices ${\displaystyle (x,\sqrt{1^2-x^2})}$, ${\displaystyle (x,0)}$, and the origin. Then ${\displaystyle \arccos \left(x\right)}$ is the angle between the positive x-axis and the ray beginning at the origin and passing through ${\displaystyle (x,\sqrt{1^2-x^2})}$. Therefore, ${\displaystyle \sin \left(\arccos \left(x\right)\right)}$ is ${\displaystyle \frac{\sqrt{1^2-x^2}}{1}}$.

Since both terms are perfect squares, factor using the difference of squares formula, ${\displaystyle a^2-b^2=(a+b)(a-b)}$ where ${\displaystyle a=1}$ and ${\displaystyle b=x}$.

Divide ${\displaystyle \sqrt{(1+x)(1-x)}}$ by ${\displaystyle 1}$.

Multiply ${\displaystyle -\sqrt{(1+x)(1-x)}\sqrt{(1+x)(1-x)}}$.

Rewrite ${\displaystyle {\sqrt{(1+x)(1-x)}}^2}$ as ${\displaystyle (1+x)(1-x)}$.

Expand ${\displaystyle (1+x)(1-x)}$ using the FOIL Method.

Simplify and combine like terms.

Apply the distributive property.

Multiply ${\displaystyle -1}$ by ${\displaystyle 1}$.

Multiply ${\displaystyle --x^2}$.