Question

Given sin x = -4/5 and x is in quadrant 3, what is the value of tan x/2

Answer 1

Answer:

tan x/2 = -2

Step-by-step explanation:

Look at the pictures.

$\sin\alpha=\dfrac{y}{r}\\\\\cos\alpha=\dfrac{x}{r}\\\\\tan\alpha=\dfrac{y}{x}\\\\\cot\alpha=\dfrac{x}{y}$

We have

$\tan\dfrac{x}{2}=\dfrac{1-\cos x}{\sin x}\\\\\sin x=-\dfrac{4}{5}\to y=-4,\ r=5\\\\\cos x=\dfrac{x}{r}$

Calculate the value of x:

$r=\sqrt{x^2+y^2}\\\\5=\sqrt{x^2+(-4)^2}\\\\5=\sqrt{x^2+16}\qquad\text{square both sides}\\\\5^2=\bigg(\sqrt{x^2+16}\bigg)^2\\\\25=x^2+16\qquad\text{subtract 16 from both sides}\\\\9=x^2\\\\x^2=9\to x=\pm\sqrt{9}\to x=-3\ or\ x=3$

In Quadrat III, cosine is negative. Therefore x = -3.

Calculate the cosine:

$\cos x=\dfrac{-3}{5}=-\dfrac{3}{5}$

Substitute to the formula of the tan x/2:

$\tan\dfrac{x}{2}=\dfrac{1-\left(-\frac{3}{5}\right)}{-\frac{4}{5}}=\dfrac{1+\frac{3}{5}}{-\frac{4}{5}}=\dfrac{\frac{8}{5}}{-\frac{4}{5}}=\dfrac{8}{5}\cdot\left(-\dfrac{5}{4}\right)\\\\=\dfrac{8\!\!\!\!\diagup^2}{5\!\!\!\!\diagup_1}\cdot\left(-\dfrac{5\!\!\!\!\diagup^1}{4\!\!\!\!\diagup_1}\right)=-2$

Answer 2

bearing in mind that, on the III Quadrant, sine as well as cosine are both negative, and that hypotenuse is never negative, so, if the sine is -4/5, the negative number must be the numerator, so sin(x) = (-4)/5.

$\bf sin(x)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-(-4)^2}=a\implies \pm\sqrt{9}=a\implies \pm 3=a \\\\\\ \stackrel{III~Quadrant}{-3=a}~\hfill cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill$

$\bf tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}\qquad \leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\[-0.35em] ~\dotfill$

$\bf tan\left( \cfrac{x}{2} \right)=\cfrac{~~\frac{-4}{5}~~}{1-\frac{3}{5}}\implies tan\left( \cfrac{x}{2} \right)=\cfrac{~~\frac{-4}{5}~~}{\frac{2}{5}}\implies tan\left( \cfrac{x}{2} \right)=\cfrac{-4}{5}\cdot \cfrac{5}{2} \\\\\\ tan\left( \cfrac{x}{2} \right)=\cfrac{-4}{2}\cdot \cfrac{5}{5}\implies tan\left( \cfrac{x}{2} \right)=-2$