Question

Please help me with this question. Will award medal.

Answer 1

well recall your pythagorean identities -> \(\textit{Pythagorean Identities}
\\ \quad \\
sin^2(\theta)+cos^2(\theta)=1
\\ \quad \\
1+cot^2(\theta)=csc^2(\theta)
\\ \quad \\
1+tan^2(\theta)=sec^2(\theta)\)


so then \(\bf sin^2(\theta)+cos^2(\theta)=1\implies sin(\theta)=\square ?\)

Answer 2

as far as where the angle lands at well

recall the range of the inverse trig functions -> \(\textit{Inverse Trigonometric Identities}
\\ \quad \\

\begin{array}{cccl}

Function&{\color{brown}{ Domain}}&{\color{blue}{ Range}}\\
\hline\\
{\color{blue}{ y}}=sin^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1&-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2}
\\ \quad \\
{\color{blue}{ y}}=cos^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1& 0 \le {\color{blue}{ y}}\le \pi
\\ \quad \\
{\color{blue}{ y}}=tan^{-1}({\color{brown}{ \theta}})&-\infty\le {\color{brown}{ \theta}} \le +\infty &-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2}
\end{array}\)

Answer 3

Would it be true; quadrants I & IV?

Answer 4

@jdoe0001

Answer 5

wel.... what did you get for \(sin(\theta)?\)