Question

Suppose θ is an angle in the fourth quadrant with cosθ=x/4. Find expressions for the other five trig functions in terms of x?

Answer 1

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

now, we know the angle is in the IV quadrant, that means, the cosine is positive and the sine is negative, or "x" is positive whilst "y" is negative.

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{x}}{\stackrel{hypotenuse}{4}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{4^2-x^2}=b\implies \stackrel{IV~quadrant}{-\sqrt{16-x^2}=b}\\\\ -------------------------------$

$\bf sin(\theta)=\cfrac{-\sqrt{16-x^2}}{4} \qquad cos(\theta)=\cfrac{x}{4} \qquad % tangent tan(\theta)=\cfrac{-\sqrt{16-x^2}}{x} \\\\\\ % cotangent cot(\theta)=\cfrac{x}{-\sqrt{16-x^2}} \qquad % cosecant csc(\theta)=\cfrac{4}{-\sqrt{16-x^2}} \qquad % secant sec(\theta)=\cfrac{4}{x}$

now, for the cotangent and the cosecant, let's rationalize the denominator,

$\bf cot(\theta)=\cfrac{x}{-\sqrt{16-x^2}}\cdot \cfrac{\sqrt{16-x^2}}{\sqrt{16-x^2}}\implies cot(\theta)=-\cfrac{x\sqrt{16-x^2}}{16-x^2} \\\\\\ csc(\theta)=\cfrac{4}{-\sqrt{16-x^2}}\cdot \cfrac{\sqrt{16-x^2}}{\sqrt{16-x^2}}\implies csc(\theta)=-\cfrac{4\sqrt{16-x^2}}{16-x^2}$