삼각치환

(a) If $\sin\theta=\dfrac{1}{20}$ and $\theta$ is between 0 and $\pi/2$, then we can draw a right triangle with angle $\theta$ that has opposite side length 1, and hypotenuse length 20. By the Pythagorean Theorem, the adjacent side has length $\sqrt{20^2-1^2}=\sqrt{399}$. So, $\cos\theta = \dfrac{\mathrm{adj}}{\mathrm{hyp}}=\dfrac{\sqrt{399}}{20}$.

We can do a quick “reasonableness" check here: $\frac{1}{20}$ is pretty close to 0, so we might expect $\theta$ to be pretty close to 0, and so $\cos \theta$ should be pretty close to 1. Indeed it is: $\dfrac{\sqrt{399}}{20}\approx \dfrac{\sqrt{400}}{20}=\dfrac{20}{20}=1$.

Alternately, we can solve this problem using identities.

$$\begin{aligned} \sin^2 \theta + \cos^2 \theta &=1\\ \left(\frac{1}{20}\right)^2+ \cos^2 \theta &=1\\ \cos\theta &= \pm\sqrt{1-\frac{1}{400}}=\pm\frac{\sqrt{399}}{20} \\ \text{Since $0 \leq \theta \leq \frac{\pi}{2}$, $\cos\theta \geq 0$, so} \\ \cos\theta &= \frac{\sqrt{399}}{20} \end{aligned}$$ (b) If $\tan\theta=7$ and $\theta$ is between 0 and $\pi/2$, then we can draw a right triangle with angle $\theta$ that has opposite side length 7 and adjacent side length 1. By the Pythagorean Theorem, the hypotenuse has length $\sqrt{7^2+1^2} = \sqrt{50}=5\sqrt{2}$. So, $\csc\theta = \dfrac{\mathrm{hyp}}{\mathrm{opp}}=\dfrac{5\sqrt{2}}{7}$.

Again, we can do a quick reasonableness check. Since 7 is much larger than 1, the triangle we're thinking of doesn't look much like the triangle in our standardized picture above: it's really quite tall, with a small base. So, the opposite side and hypotenuse are pretty close in length. Indeed, $\dfrac{5\sqrt{2}}{7}\approx 7.071$, so this dimension seems reasonable. (c) If $\sec\theta=\dfrac{\sqrt{x-1}}{2}$ and $\theta$ is between 0 and $\pi/2$, then we can draw a right triangle with angle $\theta$ that has hypotenuse length $\sqrt{x-1}$ and adjacent side length 2. By the Pythagorean Theorem, the opposite side has length $\sqrt{\sqrt{x-1}^2 - 2^2} = \sqrt{x-1-4}=\sqrt{x-5}$. So, $\tan\theta = \dfrac{\mathrm{opp}}{\mathrm{adj}}=\dfrac{\sqrt{x-5}}{2}$.

We can also solve this using identities. Note that since $\sec\theta$ exists, $\theta \neq \frac{\pi}{2}$.

$$\begin{aligned} \tan^2\theta+1&=\sec^2\theta\\ \tan^2\theta+1&=\left(\frac{\sqrt{x-1}}{2}\right)^2=\frac{x-1}{4}\\ \tan\theta &= \pm\sqrt{\frac{x-1}{4}-1} = \pm\frac{\sqrt{x-5}}{2} \\ \text{Since $0 \leq \theta < \frac{\pi}{2}$, $\tan\theta \geq 0$, so} \\ \tan\theta &= \frac{\sqrt{x-5}}{2} \end{aligned}$$