Under what condition is the identity \(\sin(\arcsin(x)) = x\) true?
For all \(x\) in the domain of arcsine: $$ x \in [-1, 1] $$ For \(|x| > 1\), the expression is undefined.
Under what condition is the identity \(\arcsin(\sin(x)) = x\) true?
Only when \(x\) lies in the principal range: $$ x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $$ If \(x\) lies outside this interval, \(x\) must first be mapped to its coterminal/reference angle in that interval.
$$ \frac{\pi}{4} $$ \(\cos\left(-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\), and \(\arccos\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}\). The principal range for arccosine is \([0, \pi]\).
$$ -\frac{\pi}{4} $$ \(\tan\left(\frac{3\pi}{4}\right) = -1\), and \(\arctan(-1) = -\frac{\pi}{4}\), which lies in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
Evaluate the exact value: $$ \cos\left(\arcsin\left(\frac{3}{5}\right)\right) $$
$$ \frac{4}{5} $$ Let \(\theta = \arcsin\left(\frac{3}{5}\right)\). Then \(\sin(\theta) = \frac{3}{5}\) in Quadrant I. Using \(\cos(\theta) = \sqrt{1 - \sin^2(\theta)}\): $$ \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5} $$
Evaluate the exact value: $$ \tan\left(\arccos\left(-\frac{5}{13}\right)\right) $$
$$ -\frac{12}{5} $$ Let \(\theta = \arccos\left(-\frac{5}{13}\right)\). \(\theta\) is in Quadrant II, so \(\cos(\theta) = -\frac{5}{13}\) and \(\sin(\theta) = \frac{12}{13}\). Thus \(\tan(\theta) = \frac{12}{-5} = -\frac{12}{5}\).
Evaluate the exact value: $$ \sin\left(\arctan\left(\frac{4}{3}\right)\right) $$
$$ \frac{4}{5} $$ Construct a right triangle with opposite side \(4\) and adjacent side \(3\). The hypotenuse is \(\sqrt{3^2 + 4^2} = 5\). Therefore, \(\sin(\theta) = \frac{4}{5}\).
Find an algebraic expression in terms of \(x\) for: $$ \cos(\arcsin(x)) \quad \text{for } x \in [-1, 1] $$
$$ \sqrt{1 - x^2} $$ Let \(\theta = \arcsin(x)\). Since \(\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), cosine is non-negative: \(\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}\).
Find an algebraic expression in terms of \(x\) for \(x > 0\): $$ \tan(\arccos(x)) $$
$$ \frac{\sqrt{1 - x^2}}{x} $$ In a reference triangle where adjacent side is \(x\) and hypotenuse is \(1\), the opposite side is \(\sqrt{1 - x^2}\). Thus \(\tan(\theta) = \frac{\sqrt{1 - x^2}}{x}\).
Find an algebraic expression in terms of \(x\) for: $$ \sin(\arctan(x)) $$
$$ \frac{x}{\sqrt{1 + x^2}} $$ In a right triangle with opposite side \(x\) and adjacent side \(1\), the hypotenuse is \(\sqrt{1 + x^2}\). Thus \(\sin(\theta) = \frac{x}{\sqrt{1 + x^2}}\).
What is the constant value of the sum for all \(x \in [-1, 1]\)? $$ \arcsin(x) + \arccos(x) $$
$$ \frac{\pi}{2} $$ In a right triangle, the two acute angles are complementary: if \(\sin(\alpha) = x\), then \(\cos\left(\frac{\pi}{2} - \alpha\right) = x\), making their sum \(\frac{\pi}{2}\).
Evaluate the exact value: $$ \cos\left(\arcsin\left(-\frac{12}{13}\right)\right) $$
$$ \frac{5}{13} $$ Let \(\theta = \arcsin\left(-\frac{12}{13}\right)\), which lies in Quadrant IV \(\left[-\frac{\pi}{2}, 0\right)\). Cosine is positive in Quadrant IV, so \(\cos(\theta) = \sqrt{1 - \left(-\frac{12}{13}\right)^2} = \frac{5}{13}\).
$$ \frac{2\pi}{3} $$ \(\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}\). The principal value of \(\arccos\left(-\frac{1}{2}\right)\) lies in Quadrant II \([0, \pi]\), which yields \(\frac{2\pi}{3}\).
