Question

If sine of x equals square root of 2 over 2, what is cos(x) and tan(x)? Explain your steps in complete sentences. (10 points)

Answer 1

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \quad % tangent tan(\theta)=\cfrac{opposite}{adjacent}\\\\ -------------------------------$

$\bf sin(x)=\cfrac{\sqrt{2}}{2}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\impliedby \textit{now, let's find the adjacent side} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{(2)^2-(\sqrt{2})^2}=a\implies \pm\sqrt{4-2}=a\implies \pm\sqrt{2}=a$

now, the adjacent side, or the x-coordinate for the angle, can be negative and positive, is positive in the I quadrant and negative in the II quadrant, with a positive y-coordinate... so it can be either, the +/- one.

$\bf cos(x)=\cfrac{\pm\sqrt{2}}{2}\qquad tan(x)=\cfrac{\frac{\sqrt{2}}{2}}{\pm\frac{\sqrt{2}}{2}}\implies tan(x)=\pm 1$