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√(c^2-a^2)=b\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm√(4^2-x^2)=b\implies \stackrel{IV~quadrant}{-√(16-x^2)=b}\\\\ -------------------------------`


`\bf sin(\theta)=\cfrac{-√(16-x^2)}{4} \qquad cos(\theta)=\cfrac{x}{4} \qquad % tangent tan(\theta)=\cfrac{-√(16-x^2)}{x} \\\\\\ % cotangent cot(\theta)=\cfrac{x}{-√(16-x^2)} \qquad % cosecant csc(\theta)=\cfrac{4}{-√(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}{-√(16-x^2)}\cdot \cfrac{√(16-x^2)}{√(16-x^2)}\implies cot(\theta)=-\cfrac{x√(16-x^2)}{16-x^2} \\\\\\ csc(\theta)=\cfrac{4}{-√(16-x^2)}\cdot \cfrac{√(16-x^2)}{√(16-x^2)}\implies csc(\theta)=-\cfrac{4√(16-x^2)}{16-x^2}`