if \\( \\tan \\alpha = - \\frac { 21 } { 20 }, 270 ^ { \\circ } < \\alpha < 360 ^ { \\circ } \\), then find…

if \\( \\tan \\alpha = - \\frac { 21 } { 20 }, 270 ^ { \\circ } < \\alpha < 360 ^ { \\circ } \\), then find the exact value of each of the following.\na. \\( \\sin \\frac { \\alpha } { 2 } \\) b. \\( \\cos \\frac { \\alpha } { 2 } \\) c. \\( \\tan \\frac { \\alpha } { 2 } \\)\na. \\( \\sin \\frac { \\alpha } { 2 } = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the exp\nb. \\( \\cos \\frac { \\alpha } { 2 } = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the exc. \\( \\tan \\frac { \\alpha } { 2 } = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the ex

if \\( \\tan \\alpha = - \\frac { 21 } { 20 }, 270 ^ { \\circ } < \\alpha < 360 ^ { \\circ } \\), then find the exact value of each of the following.\na. \\( \\sin \\frac { \\alpha } { 2 } \\) b. \\( \\cos \\frac { \\alpha } { 2 } \\) c. \\( \\tan \\frac { \\alpha } { 2 } \\)\na. \\( \\sin \\frac { \\alpha } { 2 } = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the exp\nb. \\( \\cos \\frac { \\alpha } { 2 } = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the exc. \\( \\tan \\frac { \\alpha } { 2 } = \\square \\)\n(simplify your answer, including any radicals. use integers or fractions for any numbers in the ex

Answer

Explanation:

Step1: Find (\cos\alpha)

Given (\tan\alpha =-\frac{21}{20}) and (270^{\circ}<\alpha < 360^{\circ}). Using (\tan\alpha=\frac{\sin\alpha}{\cos\alpha}) and (\sin^{2}\alpha+\cos^{2}\alpha = 1), we have (\left(-\frac{21}{20}\cos\alpha\right)^{2}+\cos^{2}\alpha=1). [ \begin{align*} \frac{441}{400}\cos^{2}\alpha+\cos^{2}\alpha&=1\ \frac{441 + 400}{400}\cos^{2}\alpha&=1\ \cos^{2}\alpha&=\frac{400}{841}\ \cos\alpha&=\frac{20}{29}\quad(\text{since in the fourth - quadrant, }\cos\alpha>0) \end{align*} ]

Step2: Use the half - angle formula for (\sin\frac{\alpha}{2})

The half - angle formula (\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{2}}). Since (135^{\circ}<\frac{\alpha}{2}<180^{\circ}) (dividing (270^{\circ}<\alpha < 360^{\circ}) by 2), (\sin\frac{\alpha}{2}>0). [ \begin{align*} \sin\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{20}{29}}{2}}\ &=\sqrt{\frac{\frac{29 - 20}{29}}{2}}\ &=\sqrt{\frac{9}{58}}\ &=\frac{3}{\sqrt{58}}=\frac{3\sqrt{58}}{58} \end{align*} ]

Step3: Use the half - angle formula for (\cos\frac{\alpha}{2})

The half - angle formula (\cos\frac{\alpha}{2}=\pm\sqrt{\frac{1+\cos\alpha}{2}}). Since (135^{\circ}<\frac{\alpha}{2}<180^{\circ}), (\cos\frac{\alpha}{2}<0). [ \begin{align*} \cos\frac{\alpha}{2}&=-\sqrt{\frac{1+\frac{20}{29}}{2}}\ &=-\sqrt{\frac{\frac{29 + 20}{29}}{2}}\ &=-\sqrt{\frac{49}{58}}\ &=-\frac{7}{\sqrt{58}}=-\frac{7\sqrt{58}}{58} \end{align*} ]

Step4: Use the half - angle formula for (\tan\frac{\alpha}{2})

The half - angle formula (\tan\frac{\alpha}{2}=\frac{\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}) or (\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}). Since (\sin\alpha=-\frac{21}{29}) (from (\tan\alpha =-\frac{21}{20}) and (\cos\alpha=\frac{20}{29})), and using (\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}) [ \begin{align*} \tan\frac{\alpha}{2}&=\frac{1-\frac{20}{29}}{-\frac{21}{29}}\ &=\frac{\frac{9}{29}}{-\frac{21}{29}}\ &=-\frac{3}{7} \end{align*} ]

Answer:

a. (\frac{3\sqrt{58}}{58}) b. (-\frac{7\sqrt{58}}{58}) c. (-\frac{3}{7})