Question

Given sinx=3/5 and x is in quadrant 2, what is the value of tanx/2

Answer 1

let's keep in mind that in the II Quadrant, sine is positive and cosine is negative, so just about the same for the opposite and adjacent sides of the angle "x", so

`\sin(x )=\cfrac{\stackrel{opposite}{3}}{\underset{hypotenuse}{5}}\hspace{5em}\textit{let's find the \underline{adjacent side}} \\\\\\ \begin{array}{llll} \textit{using the pythagorean theorem} \\\\ a^2+o^2=c^2\implies a=√(c^2 - o^2) \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{5}\\ a=adjacent\\ o=\stackrel{opposite}{3} \end{cases} \\\\\\ a=\pm√( 5^2 - 3^2) \implies a=\pm√( 16 )\implies a=\pm 4\implies \stackrel{II~Quadrant }{a=-4} \\\\[-0.35em] ~\dotfill`

`\cos(x )=\cfrac{\stackrel{adjacent}{-4}}{\underset{hypotenuse}{5}}\hspace{9em} \tan\left(\cfrac{\theta}{2}\right)=\cfrac{\sin(\theta)}{1+\cos(\theta)} \\\\\\ \tan\left(\cfrac{x}{2}\right)\implies \cfrac{(3)/(5)}{1-(4)/(5)}\implies \cfrac{ ~~ (3)/(5) ~~ }{(1)/(5)}\implies \cfrac{3}{5}\cdot \cfrac{5}{1}\implies \text{\LARGE 3}`