Question

#3 and #4 with work

Answer 1

3)


`\bf sin(\theta )=\cfrac{3}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\impliedby \textit{now let's find the \underline{adjacent} side} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm√(c^2-b^2)=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}`


`\bf \pm√(5^2-3^2)=a\implies \pm√(16)=a\implies \pm 4=a \\\\\\ \textit{now, }\theta \textit{ is in the \underline{1st quadrant}, where the adjacent is \underline{positive}} \\\\\\ \boxed{4=a}\qquad \qquad thus\qquad \qquad cos(\theta )=\cfrac{4}{5}\cfrac{\leftarrow adjacent}{\leftarrow hypotenuse}\\\\ -------------------------------\\\\ sin(2\theta)=2sin(\theta)cos(\theta)\implies 2\cdot \cfrac{4}{5}\cdot \cfrac{3}{5}\implies \cfrac{24}{25}`

4)


`\bf \theta \qquad 0\le \theta \ \textless \ 2\pi \\\\ -------------------------------\\\\ cos^2(\theta )=1\implies cos(\theta )=√(1)\implies cos(\theta )=1 \\\\\\ cos^(-1)[cos(\theta )]=cos^(-1)(1)\implies \measuredangle \theta =cos^(-1)(1)\implies \measuredangle \theta =0`