Question

Find the following EXACT:
a) sin[cos^(-1)(2/3)]
b) sin[tan^(-1)(-6)]

Answer 1

\(\bf sin\left[cos^{-1}\left(\cfrac{2}{3}\right)\right]\\ \quad \\ \quad \\
cos^{-1}\left(\cfrac{2}{3}\right)=\theta\implies cos(\theta)=\cfrac{2}{3}\implies \cfrac{\textit{adjacent side}}{\textit{hypotenus}}\implies \cfrac{a}{c}\\ \quad \\
\textit{using pythagorean theorem then}\quad c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b\\ \quad \\ \quad \\
\textit{recall that }\qquad sin(\theta)=\cfrac{b}{c}
\)

Answer 2

is there a way you could explain this in unit circle terms?

Answer 3

\(\bf sin[tan^{-1}(-6)] \\ \quad \\ \quad \\
tan^{-1}(-6)= \beta\implies tan(\beta)=-6\implies tan(\beta)=-\cfrac{6}{1}\\ \quad \\\implies \cfrac{\textit{opposite side}}{\textit{adjacent side}}\implies \cfrac{b}{a}\\ \quad \\
\textit{same as before}\\ \quad \\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}\\ \quad \\
\textit{keeping in mind that }\quad sin(\beta)=\cfrac{b}{c}\)