suppose that \\( \\sec \\alpha = \\frac { 17 } { 15 } \\) and \\( 0 ^ { \\circ } < \\alpha < 90 ^ { \\circ }…

suppose that \\( \\sec \\alpha = \\frac { 17 } { 15 } \\) and \\( 0 ^ { \\circ } < \\alpha < 90 ^ { \\circ } \\).\nfind the exact values of \\( \\cos \\frac { \\alpha } { 2 } \\) and \\( \\tan \\frac { \\alpha } { 2 } \\).\n\\( \\cos \\frac { \\alpha } { 2 } = \\)\n\\( \\tan \\frac { \\alpha } { 2 } = \\)

suppose that \\( \\sec \\alpha = \\frac { 17 } { 15 } \\) and \\( 0 ^ { \\circ } < \\alpha < 90 ^ { \\circ } \\).\nfind the exact values of \\( \\cos \\frac { \\alpha } { 2 } \\) and \\( \\tan \\frac { \\alpha } { 2 } \\).\n\\( \\cos \\frac { \\alpha } { 2 } = \\)\n\\( \\tan \\frac { \\alpha } { 2 } = \\)

Answer

Explanation:

Step1: Find (\cos\alpha)

Since (\sec\alpha=\frac{17}{15}), and (\sec\alpha=\frac{1}{\cos\alpha}), then (\cos\alpha=\frac{15}{17}).

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

The half - angle formula is (\cos\frac{\alpha}{2}=\sqrt{\frac{1 + \cos\alpha}{2}}). Substitute (\cos\alpha=\frac{15}{17}) into the formula: [ \begin{align*} \cos\frac{\alpha}{2}&=\sqrt{\frac{1+\frac{15}{17}}{2}}\ &=\sqrt{\frac{\frac{17 + 15}{17}}{2}}\ &=\sqrt{\frac{32}{17\times2}}\ &=\sqrt{\frac{16}{17}}\ &=\frac{4}{\sqrt{17}}\ &=\frac{4\sqrt{17}}{17} \end{align*} ]

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

The half - angle formula is (\tan\frac{\alpha}{2}=\sqrt{\frac{1-\cos\alpha}{1 + \cos\alpha}}). Substitute (\cos\alpha=\frac{15}{17}) into the formula: [ \begin{align*} \tan\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{15}{17}}{1+\frac{15}{17}}}\ &=\sqrt{\frac{\frac{17-15}{17}}{\frac{17 + 15}{17}}}\ &=\sqrt{\frac{2}{32}}\ &=\sqrt{\frac{1}{16}}\ &=\frac{1}{4} \end{align*} ]

Answer:

(\cos\frac{\alpha}{2}=\frac{4\sqrt{17}}{17}), (\tan\frac{\alpha}{2}=\frac{1}{4})