Write cos(arcsin3x +arccos x) as an algebraic expression of x that does not involve trigonometric functions.

Answer:[tex]x \sqrt{1-9x^2} -3x \sqrt{1-x^2}[/tex]Step-by-step explanation:Since we have to analyze the cosine of an addition of angles, let's recall that cos of an addition is:[tex]cos (\alpha +\beta ) = cos(\alpha ) cos(\beta) - sin(\alpha) sin(\beta)[/tex]Let's also use those Greek letters [tex]\alpha[/tex] and [tex]\alpha[/tex] to represent the inverse functions we are dealing with:[tex]\alpha =arcsin(3x)[/tex] and therefore, [tex]sin(\alpha ) = 3x[/tex][tex]\beta =arccos(x)[/tex] and therefore [tex]cos(\beta ) = x[/tex]Now use these identities to re-write the equation for the cos of an addition given above:[tex]cos (\alpha +\beta ) = cos(\alpha ) cos(\beta) - sin(\alpha) sin(\beta)\\cos (\alpha +\beta ) = cos(\alpha ) x - 3x sin(\beta )[/tex]Now recall that the sine of an angle can be written in terms of the cos of the same angle via the Pythagorean identity: [tex]sin(\beta ) = \sqrt{1-cos^2(\beta) }[/tex]In our case, and with our particular values, this gives:[tex]sin(\beta ) = \sqrt{1-cos^2(\beta)} = \sqrt{1-x^2}[/tex]Similarly we can write the cos of an angle in terms of the sine of the same angle:[tex]cos(\alpha ) = \sqrt{1-sin^2(\alpha) }[/tex]In ours case, using our values, this gives:[tex]cos(\alpha ) = \sqrt{1-sin^2(\alpha) }= \sqrt{1-(3x)^2}= \sqrt{1-9x^2}[/tex]So replacing these in the cos of an addition of angles formula, we obtain that the given expression equals:[tex]x \sqrt{1-9x^2} -3x \sqrt{1-x^2}[/tex]