Certainly! Let's find an algebraic expression for [tex]\(\cos(\arcsin(3x) + \arccos(x))\)[/tex] that does not involve trigonometric functions.
To solve this, we can use the following trigonometric identities:
1. [tex]\(\arcsin(y) + \arccos(y) = \frac{\pi}{2}\)[/tex] for any [tex]\(y\)[/tex] within the domain of both arctangent functions,
2. [tex]\(\cos(A + B) = \cos A \cos B - \sin A \sin B\)[/tex].
Given:
[tex]\[ A = \arcsin(3x) \][/tex]
[tex]\[ B = \arccos(x) \][/tex]
We need to find:
[tex]\[ \cos(\arcsin(3x) + \arccos(x)) \][/tex]
From the identity [tex]\(\cos(A + B) = \cos A \cos B - \sin A \sin B\)[/tex]:
[tex]\[ \cos(\arcsin(3x) + \arccos(x)) = \cos(\arcsin(3x)) \cos(\arccos(x)) - \sin(\arcsin(3x)) \sin(\arccos(x)) \][/tex]
Now, evaluate each trigonometric function:
1. [tex]\(\cos(\arcsin(3x))\)[/tex]:
[tex]\[ \sin(\arcsin(3x)) = 3x \][/tex]
We know that [tex]\(\sin^2 A + \cos^2 A = 1\)[/tex], so,
[tex]\[ \cos(\arcsin(3x)) = \sqrt{1 - \sin^2(\arcsin(3x))} = \sqrt{1 - (3x)^2} = \sqrt{1 - 9x^2} \][/tex]
2. [tex]\(\cos(\arccos(x)) = x\)[/tex] (by definition).
3. [tex]\(\sin(\arccos(x))\)[/tex]:
[tex]\[ \cos(\arccos(x)) = x \][/tex]
Similarly, [tex]\(\sin^2 B + \cos^2 B = 1\)[/tex], so,
[tex]\[ \sin(\arccos(x)) = \sqrt{1 - \cos^2(\arccos(x))} = \sqrt{1 - x^2} \][/tex]
Put these back into our main expression:
[tex]\[ \cos(\arcsin(3x)) \cos(\arccos(x)) - \sin(\arcsin(3x)) \sin(\arccos(x)) \][/tex]
[tex]\[ = (\sqrt{1 - 9x^2})(x) - (3x)(\sqrt{1 - x^2}) \][/tex]
[tex]\[ = x \sqrt{1 - 9x^2} - 3x \sqrt{1 - x^2} \][/tex]
Therefore, the algebraic expression for [tex]\(\cos(\arcsin(3x) + \arccos(x))\)[/tex] is:
[tex]\[ \boxed{x \sqrt{1 - 9x^2} - 3x \sqrt{1 - x^2} } \][/tex]
