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Prove that cos^2A+cos^2(A+120)+cos^2(A-120)=3/2 |
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Answer» Cos5θ=16〖Sin〗^5 θ-20〖Sin〗^3 θ+5cosθ {tex}{\\cos ^2}2{\\rm{A}} + {\\cos ^2}2\\left( {{\\rm{A}} + {{120}^ \\circ }} \\right) + {\\cos ^2}2\\left( {{\\rm{A}} - {{120}^ \\circ }} \\right) = {3 \\over 2}{/tex}We know that{tex}{\\cos ^2}{\\rm{A}} = {{1 + \\cos 2{\\rm{A}}} \\over 2}{/tex} ..............(i){tex}{\\cos ^2}2\\left( {{\\rm{A}} + {{120}^ \\circ }} \\right) = {{1 + \\cos \\left( {{\\rm{2A}} + {{240}^ \\circ }} \\right)} \\over 2}{/tex} ..............(ii)And {tex}{\\cos ^2}2\\left( {{\\rm{A}} - {{120}^ \\circ }} \\right) = {{1 + \\cos \\left( {{\\rm{2A}} - {{240}^ \\circ }} \\right)} \\over 2}{/tex}..............(iii)Adding all these three equations,{tex}{\\cos ^2}2{\\rm{A}} + {\\cos ^2}2\\left( {{\\rm{A}} + {{120}^ \\circ }} \\right) + {\\cos ^2}2\\left( {{\\rm{A}} - {{120}^ \\circ }} \\right){/tex}\xa0= {tex}{3 \\over 2} + {{1 + \\cos 2{\\rm{A}}} \\over 2} + {{1 + \\cos \\left( {{\\rm{2A + }}{{240}^ \\circ }} \\right)} \\over 2} + {{1 + \\cos \\left( {{\\rm{2A}} - {{240}^ \\circ }} \\right)} \\over 2}{/tex}= {tex}{{3 + 1 + \\cos 2{\\rm{A + }}1 + \\cos \\left( {{\\rm{2A + }}{{240}^ \\circ }} \\right) + 1 + \\cos \\left( {{\\rm{2A}} - {{240}^ \\circ }} \\right)} \\over 2}{/tex}= {tex}{{3 + 3 + \\cos 2{\\rm{A}} + \\cos \\left( {{\\rm{2A + }}{{240}^ \\circ }} \\right) + \\cos \\left( {{\\rm{2A}} - {{240}^ \\circ }} \\right)} \\over 2}{/tex}= {tex}{{3 + 3 + \\cos 2{\\rm{A}} + 2\\cos 2{\\rm{A}}.\\cos {{240}^ \\circ }} \\over 2}{/tex}= {tex}{{3 + 3 + \\cos 2{\\rm{A}} - \\cos 2{\\rm{A}}} \\over 2}{/tex}= {tex}{6 \\over 2}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}{\\cos ^2}2{\\rm{A}} + {\\cos ^2}2\\left( {{\\rm{A}} + {{120}^ \\circ }} \\right) + {\\cos ^2}2\\left( {{\\rm{A}} - {{120}^ \\circ }} \\right) = {3 \\over 2}{/tex} |
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