Hello there. To solve this question, we'll have to remember some properties about inverse trigonometric functions.
Given the expression:
[tex]\sin (\sin ^{-1}(x)-\cos ^{-1}(x))[/tex]
We want to express it as an algebraic function of x.
For this, imagine the following triangle:
Now, we find the missing leg of the triangle applying Pythagoras theorem:
[tex]\begin{gathered} 1^2=x^2+?^2 \\ 1-x^2=?^2 \\ ?^{}=\sqrt[]{1-x^2} \end{gathered}[/tex]
Now, finding the sine and cosine of the angles alpha and beta, we get:
First, remember the sine of an angle is equal to the ratio between the opposite side to the angle and the hypotenuse of the triangle. The cosine of the same angle is equal to the ratio between the adjacent side to the angle and the hypotenuse of the triangle.
Therefore, we have:
[tex]\begin{gathered} \sin (\alpha)=\frac{x}{1}=x \\ \cos (\beta)=\frac{x}{1}=x \\ \sin (\beta)=\frac{\sqrt[]{1-x^2}}{1}=\sqrt[]{1-x^2} \\ \cos (\alpha)=\frac{\sqrt[]{1-x^2}}{1}=\sqrt[]{1-x^2} \end{gathered}[/tex]
This means that:
[tex]\begin{gathered} \alpha=\sin ^{-1}(x) \\ \beta=\cos ^{-1}(x) \end{gathered}[/tex]
Now, the expression we had earlier turns into:
[tex]\sin (\alpha-\beta)[/tex]
For this, we'll use the angle sum formula:
[tex]\sin (u-v)=\sin (u)\cos (v)-\sin (v)\cos (u)[/tex]
Which gives us:
[tex]\sin (\alpha)\cos (\beta)-\sin (\beta)\cos (\alpha)_{}[/tex]
Plugging the results we got earlier, this is simply:
[tex]x\cdot x-\sqrt[]{1-x^2}\cdot\sqrt[]{1-x^2}[/tex]
As x > 0, because we're using it as a triangle side (but it could be negative considering inverse sine and cosine as functions), we get:
[tex]\begin{gathered} x^2-(1-x^2) \\ x^2-1+x^2 \\ 2x^2-1 \end{gathered}[/tex]
