suppose that \\( \\cot \\alpha=-3 \\) and \\( 90^{\\circ}<\\alpha<180^{\\circ} \\).\nfind the exact values…

suppose that \\( \\cot \\alpha=-3 \\) and \\( 90^{\\circ}<\\alpha<180^{\\circ} \\).\nfind the exact values of \\( \\sin \\frac{\\alpha}{2} \\) and \\( \\tan \\frac{\\alpha}{2} \\).\n\\\\begin{array}{l}\\sin \\frac{\\alpha}{2}= \\\\ \\tan \\frac{\\alpha}{2}=\\end{array}\\

suppose that \\( \\cot \\alpha=-3 \\) and \\( 90^{\\circ}<\\alpha<180^{\\circ} \\).\nfind the exact values of \\( \\sin \\frac{\\alpha}{2} \\) and \\( \\tan \\frac{\\alpha}{2} \\).\n\\\\begin{array}{l}\\sin \\frac{\\alpha}{2}= \\\\ \\tan \\frac{\\alpha}{2}=\\end{array}\\

Answer

Explanation:

Step1: Find (\sin\alpha) and (\cos\alpha)

Given (\cot\alpha=-3=\frac{\cos\alpha}{\sin\alpha}), and (\sin^{2}\alpha+\cos^{2}\alpha = 1). Substitute (\cos\alpha=-3\sin\alpha) into (\sin^{2}\alpha+\cos^{2}\alpha = 1), we get (\sin^{2}\alpha+(-3\sin\alpha)^{2}=1), i.e., (\sin^{2}\alpha + 9\sin^{2}\alpha=1), (10\sin^{2}\alpha=1), (\sin\alpha=\pm\frac{1}{\sqrt{10}}). Since (90^{\circ}<\alpha<180^{\circ}), (\sin\alpha=\frac{1}{\sqrt{10}}), then (\cos\alpha=-3\times\frac{1}{\sqrt{10}}=-\frac{3}{\sqrt{10}}).

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

The half - angle formula (\sin\frac{\alpha}{2}=\sqrt{\frac{1 - \cos\alpha}{2}}). Substitute (\cos\alpha=-\frac{3}{\sqrt{10}}) into it: (\sin\frac{\alpha}{2}=\sqrt{\frac{1+\frac{3}{\sqrt{10}}}{2}}=\sqrt{\frac{\sqrt{10} + 3}{2\sqrt{10}}}=\sqrt{\frac{10 + 3\sqrt{10}}{20}}=\frac{\sqrt{10 + 3\sqrt{10}}}{2\sqrt{5}}=\frac{\sqrt{20 + 6\sqrt{10}}}{10}). Another way: Since (45^{\circ}<\frac{\alpha}{2}<90^{\circ}), (\sin\frac{\alpha}{2}>0). We know (\cos\alpha=-\frac{3}{\sqrt{10}}), then (\sin\frac{\alpha}{2}=\sqrt{\frac{1+\frac{3}{\sqrt{10}}}{2}}=\frac{\sqrt{\sqrt{10}+3}}{\sqrt{2\sqrt{10}}}=\frac{\sqrt{10 + 3\sqrt{10}}}{\sqrt{20}}=\frac{\sqrt{10+3\sqrt{10}}}{2\sqrt{5}}=\frac{\sqrt{50 + 15\sqrt{10}}}{10}). Simplify (\sin\frac{\alpha}{2}=\sqrt{\frac{1-\cos\alpha}{2}}=\sqrt{\frac{1+\frac{3}{\sqrt{10}}}{2}}=\sqrt{\frac{\sqrt{10}+3}{2\sqrt{10}}}=\frac{\sqrt{10 + 3\sqrt{10}}}{2\sqrt{5}}=\frac{\sqrt{50+15\sqrt{10}}}{10}=\frac{\sqrt{10}+3}{2\sqrt{5}}=\frac{\sqrt{50}+3\sqrt{5}}{10}=\frac{\sqrt{10}+3}{2\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{50}+3\sqrt{5}}{10}=\frac{5\sqrt{2}+3\sqrt{5}}{10}) (wrong, re - calculate). Correct: (\sin\frac{\alpha}{2}=\sqrt{\frac{1-\cos\alpha}{2}}), (\cos\alpha =-\frac{3}{\sqrt{10}}), (\sin\frac{\alpha}{2}=\sqrt{\frac{1+\frac{3}{\sqrt{10}}}{2}}=\sqrt{\frac{\sqrt{10}+3}{2\sqrt{10}}}=\frac{\sqrt{10 + 3\sqrt{10}}}{2\sqrt{5}}=\frac{\sqrt{50+15\sqrt{10}}}{10}) (complex, use another formula). Since (\cot\alpha=-3), (\sin\alpha=\frac{1}{\sqrt{10}}), (\cos\alpha=-\frac{3}{\sqrt{10}}). (\sin\frac{\alpha}{2}=\sqrt{\frac{1-\cos\alpha}{2}}=\sqrt{\frac{1+\frac{3}{\sqrt{10}}}{2}}=\frac{\sqrt{\sqrt{10}+3}}{\sqrt{2\sqrt{10}}}=\frac{\sqrt{10 + 3\sqrt{10}}}{\sqrt{20}}=\frac{\sqrt{10+3\sqrt{10}}}{2\sqrt{5}}=\frac{\sqrt{50 + 15\sqrt{10}}}{10}) (not simple). Use (\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}) (a better formula for (\tan\frac{\alpha}{2})).

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

The half - angle formula (\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}). Substitute (\sin\alpha=\frac{1}{\sqrt{10}}) and (\cos\alpha=-\frac{3}{\sqrt{10}}) into it: (\tan\frac{\alpha}{2}=\frac{1+\frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}}=\sqrt{10}+3).

Answer:

(\sin\frac{\alpha}{2}=\frac{\sqrt{10 + 3\sqrt{10}}}{2\sqrt{5}}) (or simplified as (\frac{\sqrt{50+15\sqrt{10}}}{10})), (\tan\frac{\alpha}{2}=\sqrt{10}+3)