Answer:
[tex]2x^2-1[/tex]
Explanation:
Given the below function;
[tex]\sin (\sin ^{-1}(x)+\cos ^{-1}(x))[/tex]
Let;
[tex]\begin{gathered} \sin ^{-1}\mleft(x\mright)=a \\ \cos ^{-1}\mleft(x\mright)=b \end{gathered}[/tex]
So we'll have;
[tex]\begin{gathered} \sin a=\cos b=x \\ \therefore\cos a=\sin b=\pm\sqrt[]{1-x^2} \end{gathered}[/tex]
We can now rewrite the given expression as;
[tex]\begin{gathered} \sin (a+b)=\sin a\cos b+\cos a\sin b \\ =x\cdot x+(\pm\sqrt[]{1-x^2})\cdot(\pm\sqrt[]{1-x^2}) \\ =x^2\pm(1-x^2) \\ So\text{ we'll have;} \\ \Rightarrow x^2+(1-x^2) \\ x^2+1-x^2=1 \\ Or \\ \Rightarrow x^2-(1-x^2) \\ x^2-1+x^2=2x^2-1 \end{gathered}[/tex]
We can see from the above that the algebraic function of x is 2x^2 - 1
