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{√(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√(c^2-a^2)=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}`


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