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

`\tan(\theta)/(2)=-2`

Explanation:

`Since\ \theta\ in\ Quadrant\ III\\\Let\ \theta=180+\phi\\\\\cos\theta=-(3)/(5)\\\\\cos(180+\phi)=-(3)/(5)\\\\-\cos\phi=-(3)/(5)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ as\ \cos(180+x)=-\cos x\\\\\cos\phi=(3)/(5)\\\\2\cos^2(\phi)/(2)-1=(3)/(5)\ \ \ \ \ \ \ \ \ \ as\ \cos x=2\cos^2(x)/(2)-1\\\\2\cos^2(\phi)/(2)=1+(3)/(5)=(8)/(5)\\\\\cos^2(\phi)/(2)=(8)/(5* 2)\\\\\cos^2(\phi)/(2)=(4)/(5)`

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