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]
