Question

Given the function value of the acute angle find the other five trigonometric function values cos a= Sqrt 7/7

Answer 1

Well, we know the angle is acute, that means is less than 90°, meaning is in the first quadrant, that means the "x" or adjacent side as well as the "y" or opposite side, are both positive.

$\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\\\\ -------------------------------\\\\ cos(a)=\cfrac{\sqrt{7}}{7}\cfrac{\leftarrow adjacent}{\leftarrow hypotenuse}\quad \textit{now, let's find the \underline{opposite}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}$

$\bf \pm\sqrt{7^2-(\sqrt{7})^2}=b\implies \pm\sqrt{49-7}=b\implies \pm\sqrt{42}=b \\\\\\ \textit{now, which is it, the + or -? well, we're in the 1st quadrant, is }\sqrt{42}$

$\bf \textit{now that you know all three sides, just plug them in} \\\\\\ sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad % tangent tan(\theta)=\cfrac{opposite}{adjacent} \\\\\\ % cotangent cot(\theta)=\cfrac{adjacent}{opposite} \qquad % cosecant csc(\theta)=\cfrac{hypotenuse}{opposite} \quad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}$