Question

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

Answer 1

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 √(c^2-b^2)=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm√(5^2-(-4)^2)=a\implies \pm√(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{~~(-4)/(5)~~}{1-(3)/(5)}\implies tan\left( \cfrac{x}{2} \right)=\cfrac{~~(-4)/(5)~~}{(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`

Answer 2

tan x/2 = -2

Explanation:

Look at the pictures.

`\sin\alpha=(y)/(r)\\\\\cos\alpha=(x)/(r)\\\\\tan\alpha=(y)/(x)\\\\\cot\alpha=(x)/(y)`

We have

`\tan(x)/(2)=(1-\cos x)/(\sin x)\\\\\sin x=-(4)/(5)\to y=-4,\ r=5\\\\\cos x=(x)/(r)`

Calculate the value of x:

`r=√(x^2+y^2)\\\\5=√(x^2+(-4)^2)\\\\5=√(x^2+16)\qquad\text{square both sides}\\\\5^2=\bigg(√(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√(9)\to x=-3\ or\ x=3`

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

Calculate the cosine:

`\cos x=(-3)/(5)=-(3)/(5)`

Substitute to the formula of the tan x/2:

`\tan(x)/(2)=(1-\left(-(3)/(5)\right))/(-(4)/(5))=(1+(3)/(5))/(-(4)/(5))=((8)/(5))/(-(4)/(5))=(8)/(5)\cdot\left(-(5)/(4)\right)\\\\=(8\!\!\!\!\diagup^2)/(5\!\!\!\!\diagup_1)\cdot\left(-(5\!\!\!\!\diagup^1)/(4\!\!\!\!\diagup_1)\right)=-2`