삼각치환

(a) Let $\theta = \arccos \left(\frac{x}{2}\right)$. That is, $\cos(\theta) = \frac{x}{2}$, and $0 \leq \theta \leq \pi$. Then we can draw the corresponding right triangle with angle $\theta$ with adjacent side of signed length $x$ (we note that if $\theta > \frac{\pi}{2}$, then $x$ is negative) and hypotenuse of length $2$. By the Pythagorean Theorem, the opposite side of the triangle has length $\sqrt{4-x^2}$.

So,

$$\sin\left(\arccos \left(\frac{x}{2}\right)\right)=\sin \theta = \frac{\mathrm{opp}}{\mathrm{hyp}} = \frac{\sqrt{4-x^2}}{2}$$ (b) Let $\theta = \arctan \left(\frac{1}{\sqrt{3}}\right)$. That is, $\tan(\theta) = \frac{1}{\sqrt{3}}$, and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

• Solution 1: Then $\theta = \dfrac{\pi}{6}$, so $\sin\theta = \dfrac{1}{2}$.
• Solution 2: Then we can draw the corresponding right triangle with angle $\theta$ with opposite side of length $1$ and adjacent side of length $\sqrt{3}$. By the Pythagorean Theorem, the hypotenuse of the triangle has length $\sqrt{\sqrt{3}^2+1^2}=2$.

So,

$$\sin\left(\arctan \left(\frac{1}{\sqrt{3}}\right)\right)=\sin \theta = \frac{\mathrm{opp}}{\mathrm{hyp}} = \frac{1}{2}$$ (c) Let $\theta = \arcsin \left(\sqrt{x}\right)$. That is, $\sin(\theta) = \sqrt{x}$, and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$. Then we can draw the corresponding right triangle with angle $\theta$ with opposite side of length $\sqrt{x}$ and hypotenuse of length $1$. By the Pythagorean Theorem, the adjacent side of the triangle has length $\sqrt{1-x}$.

So,

$$\sec\left(\arcsin \left(\sqrt{x}\right)\right)=\sec \theta = \frac{\mathrm{hyp}}{\mathrm{adj}} = \frac{1}{\sqrt{1-x}}$$