Question

If θ is an angle in standard position that terminates in Quadrant III such that cosθ = -3/5, then tan =θ/2 _____.

Answer 1

Answer:

$\tan\frac{\theta}{2}=-2$

Step-by-step explanation:

$Since\ \theta\ in\ Quadrant\ III\\\Let\ \theta=180+\phi\\\\\cos\theta=-\frac{3}{5}\\\\\cos(180+\phi)=-\frac{3}{5}\\\\-\cos\phi=-\frac{3}{5}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ as\ \cos(180+x)=-\cos x\\\\\cos\phi=\frac{3}{5}\\\\2\cos^2\frac{\phi}{2}-1=\frac{3}{5}\ \ \ \ \ \ \ \ \ \ as\ \cos x=2\cos^2\frac{x}{2}-1\\\\2\cos^2\frac{\phi}{2}=1+\frac{3}{5}=\frac{8}{5}\\\\\cos^2\frac{\phi}{2}=\frac{8}{5\times 2}\\\\\cos^2\frac{\phi}{2}=\frac{4}{5}$

$\cos\frac{\phi}{2}=\sqrt{\frac{4}{5}}\ \ \ \ \ positive\ sign\ is\ taken\ as\ \frac{\phi}{2}\ is\ in\ first\ quadrant\ and\ \cos\ is\ positive\ there.\\\\\sin\frac{\phi}{2}=\sqrt{1-\cos^2\frac{\phi}{2}}=\sqrt{1-\frac{4}{5}}=\sqrt\frac{1}{5}\\\\\theta=180+\phi\\\\divide\ by\ 2\\\\\frac{\theta}{2}=\frac{180+\phi}{2}\\\\\frac{\theta}{2}=90+\frac{\phi}{2}\\\\\tan\frac{\theta}{2}=\tan(90+\frac{\phi}{2})\\\\\tan\frac{\theta}{2}=-\cot\frac{\phi}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ as\ \tan(90+x)=-\cot x$$\tan\frac{\theta}{2}=-\frac{\cos\frac{\phi}{2}}{\sin\frac{\phi}{2}}\\\\\tan\frac{\theta}{2}=-\frac{\sqrt{\frac{4}{5}}}{\sqrt{\frac{1}{5}}}\\\\\tan\frac{\theta}{2}=-\sqrt{\frac{4}{1}}\\\\\tan\frac{\theta}{2}=-2$