Question

If the sin=-3/5 , where cos>0 then what are the values of the remaining trig functions

Answer 1

now, keeping in mind that, sine is opposite/hypotenuse, and that the hypotenuse is just a radius unit, and therefore is never negative, then, in the fraction of -3/5, the negative values has to be the numerator, not the hypotenuse.

something else to keep in mind is that, cosine > 0, meaning is positive, that only happens in the I and IV quadrants.

since we know the sine is negative, and the cosine is positive, the only place that occurs is on the IV quadrant, so then θ is in the IV quadrant.


`\bf sin(\theta )=\cfrac{\stackrel{opposite}{-3}}{\stackrel{hypotenuse}{5}}\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} \\\\\\ \pm√(5^2-(-3)^2)=a\implies \pm√(25-9)=a\implies \pm√(16)=a \\\\\\ \pm 4=a\implies \stackrel{IV~quadrant}{4=a}`


`\bf cos(\theta)=\cfrac{adjacent}{hypotenuse} \qquad % tangent tan(\theta)=\cfrac{opposite}{adjacent} \\\\\\ % cotangent cot(\theta)=\cfrac{adjacent}{opposite} \qquad % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} \quad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}\\\\ -------------------------------\\\\ cos(\theta)=\cfrac{4}{5}\qquad tan(\theta)=\cfrac{-3}{4}\qquad cot(\theta)=\cfrac{4}{-3} \\\\\\ csc(\theta)=\cfrac{5}{-3}\qquad sec(\theta)=\cfrac{5}{4}`