Question

Hey I'm not too sure about this. What would the exact trig value be? Thanks.

Answer 1

well, let's bear in mind that the hypotenuse is never negative, since it's just a radius unit.

now, π < θ < 3π/2, is another way of saying, the angle θ is in the III Quadrant, where as you already know, the sine and cosine as well as opposite and adjacent sides are both negative.

`\bf sin(\theta )=\cfrac{\stackrel{opposite}{-1}}{\stackrel{hypotenuse}{5}}\qquad \impliedby \textit{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-(-1)^2)=a\implies \pm√(24)=a\implies \stackrel{III~Quadrant}{-√(24)=a}\implies -2√(6)=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{-2√(6)}}{\stackrel{hypotenuse}{5}}~\hfill`