Question

#22 and #23 with work please

Answer 1

22)


`\bf \textit{let's say that }sin^(-1)\left( (2)/(3) \right)=\theta \textit{ this means }sin(\theta )=\cfrac{2}{3}\qquad so \\\\\\ sin\left[ sin^(-1)\left( (2)/(3) \right) \right]\implies sin[\theta ]\implies \cfrac{2}{3}`

23)


`\bf sec^(-1)\left( (5)/(2) \right)=\theta \textit{ this means }sec(\theta )=\cfrac{5}{2}\cfrac{\leftarrow hypotenuse}{\leftarrow adjacent} \\\\\\ \textit{let's find the \underline{opposite side} then}`


`\bf \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm√(c^2-a^2)=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm√(5^2-2^2)=b\implies \boxed{\pm √(21)=b}\\\\ -------------------------------\\\\ tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{\pm√(21)}{2}`

again, we dunno what quadrant the angle is at, thus the +/- are both valid.