Respuesta :
let us take a peek of the double-angle identities, check your book on them
[tex]\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 \\\\\\ [/tex]
[tex]\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}}[/tex]
[tex]\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 \\\\\\ [/tex]
[tex]\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}}[/tex]
The answer is 0.54. I used the good old Half-Angle Formula to get this answer.
