question help: video\nquestion 39\nevaluate: csc⁻¹(2)\nquestion 40\nfind arccos(cos 310°).\narccos(cos…

Question 39

Explanation:

Step1: Recall csc inverse definition

Let $y = \csc^{-1}(2)$. Then $\csc(y)=2$, and since $\csc(y)=\frac{1}{\sin(y)}$, we have $\sin(y)=\frac{1}{2}$. We know that the range of $y = \csc^{-1}(x)$ is $[-\frac{\pi}{2},0)\cup(0,\frac{\pi}{2}]$ and for $\sin(y)=\frac{1}{2}$ in this range, $y=\frac{\pi}{6}$ (in radians) or $30^{\circ}$.

Answer:

$30^{\circ}$ (or $\frac{\pi}{6}$ radians)

Question 40

Explanation:

Step1: Recall arccos property

The function $y = \arccos(x)$ has a range of $[0,180^{\circ}]$. We know that $\cos(310^{\circ})=\cos(360^{\circ} - 50^{\circ})=\cos(50^{\circ})$ (using the property $\cos(360^{\circ}-\alpha)=\cos(\alpha)$). Since $50^{\circ}\in[0,180^{\circ}]$, $\arccos(\cos(310^{\circ})) = 50^{\circ}$.

Answer:

$50^{\circ}$

Question 41

Explanation:

Step1: Let $\theta=\arctan(\frac{5}{17})$

Then $\tan(\theta)=\frac{5}{17}$, which means in a right - triangle, if the opposite side to angle $\theta$ is $a = 5$ and the adjacent side is $b = 17$.

Step2: Find the hypotenuse

By the Pythagorean theorem $c=\sqrt{a^{2}+b^{2}}=\sqrt{5^{2}+17^{2}}=\sqrt{25 + 289}=\sqrt{314}$.

Step3: Find secant value

Since $\sec(\theta)=\frac{c}{b}$, and $c=\sqrt{314}$, $b = 17$, then $\sec(\arctan(\frac{5}{17}))=\frac{\sqrt{314}}{17}$.

Answer:

$\frac{\sqrt{314}}{17}$