Question

Find the exact values of the six trigonometric functions of given the point (-4, 7) on the terminal side of in standard position.

Answer 1

`\bf (\stackrel{a}{-4}~,~\stackrel{b}{7})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies c=√(a^2+b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=√((-4)^2+7^2)\implies c=√(65)`

keeping in mind that, even though the square root would give us two roots, the positive and negative ones, the hypotenuse is however only a radius distance, and therefore is never negative.


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


`\bf sin(\theta )=\cfrac{7}{√(65)}\quad \stackrel{rationalized}{\implies }\quad\cfrac{7√(65)}{65} \\\\\\ cos(\theta )=\cfrac{-4}{√(65)}\quad \stackrel{rationalized}{\implies }\quad \cfrac{-4√(65)}{65} \\\\\\ tan(\theta )=-\cfrac{7}{4}\qquad\qquad \qquad cot(\theta )=-\cfrac{4}{7} \\\\\\ csc(\theta )=\cfrac{√(65)}{7}\qquad\qquad \qquad sec(\theta )=-\cfrac{√(65)}{4}`