what is a possible value of sin theta when cos 2 theta=0.42?
b.0.54
c.0.46
d.0.84
let us take a peek of the double-angle identities, check your book on them
\(\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
\boxed{1-2sin^2(\theta)}\leftarrow
\begin{array}{llll}
\textit{let us use}\\
\textit{this one}
\end{array}\\
2cos^2(\theta)-1
\end{cases}
\\ \quad \\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
-----------------------------\\\\
cos(2\theta)=1-2sin^2(\theta)\qquad \textit{now, we know that }cos(2\theta)=0.42
\\\\\\
thus
\\\\\\\)
\(\bf 0.42=1-2sin^2(\theta)\implies 2sin^2(\theta)=1-0.42
\\\\\\
sin^2(\theta)=\cfrac{1-0.42}{2}\implies sin(\theta)=\sqrt{\cfrac{1-0.42}{2}}\)
The answer is 0.54. I used the good old Half-Angle Formula to get this answer.