3)
[tex]\bf sin(\theta )=\cfrac{3}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\impliedby \textit{now let's find the \underline{adjacent} side}
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\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}[/tex]
[tex]\bf \pm\sqrt{5^2-3^2}=a\implies \pm\sqrt{16}=a\implies \pm 4=a
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\textit{now, }\theta \textit{ is in the \underline{1st quadrant}, where the adjacent is \underline{positive}}
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\boxed{4=a}\qquad \qquad thus\qquad \qquad cos(\theta )=\cfrac{4}{5}\cfrac{\leftarrow adjacent}{\leftarrow hypotenuse}\\\\
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sin(2\theta)=2sin(\theta)cos(\theta)\implies 2\cdot \cfrac{4}{5}\cdot \cfrac{3}{5}\implies \cfrac{24}{25}[/tex]
4)
[tex]\bf \theta \qquad 0\le \theta \ \textless \ 2\pi \\\\
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cos^2(\theta )=1\implies cos(\theta )=\sqrt{1}\implies cos(\theta )=1
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cos^{-1}[cos(\theta )]=cos^{-1}(1)\implies \measuredangle \theta =cos^{-1}(1)\implies \measuredangle \theta =0[/tex]
