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| 1. |
Prove that cos4π/8 + cos43π/8 + cos45π/8 + cos4\xa07π/8 = 3/2 |
| Answer» Ans.\xa0\\(cos^4{\\pi \\over 8}+cos^4{3\\pi \\over 8}+cos^4{4\\pi \\over 8}+cos^4{7\\pi \\over 8} ={3\\over 2}\\)Taking LHS,\\(cos^4{\\pi \\over 8}+cos^4{3\\pi \\over 8}+cos^4{4\\pi \\over 8}+cos^4{7\\pi \\over 8}\\)=>\xa0\\(cos^4{\\pi \\over 8}+cos^4{3\\pi \\over 8}+cos^4[{{\\pi} -{3\\pi \\over 8}}]+cos^4[{{\\pi - {\\pi \\over 8}}}]\\)=>\xa0\\(cos^4{\\pi \\over 8}+cos^4{3\\pi \\over 8}+({-cos{3\\pi \\over 8}})^4+({-cos{\\pi \\over 8}})^4\\)=>\xa0\\(2[cos^4{\\pi \\over 8}+cos^4{3\\pi \\over 8}]\\)=>\xa0\\(2[cos^4{\\pi \\over 8}+cos^4({\\pi \\over 2}-{\\pi \\over 8})]\\)=>\xa0\\(2[cos^4{\\pi \\over 8}+sin^4{\\pi \\over 8}]\\)\\(=> 2[(cos^2{\\pi \\over 8}+sin^2{\\pi \\over 8})^2 -2 cos^2{\\pi \\over 8}.sin^2{\\pi \\over 8}]\\)\\(=> 2[(1)^2 -{1\\over 2}(2 cos{\\pi \\over 8}.sin{\\pi \\over 8})^2]\\)\\(=> 2[1 -{1\\over 2}(sin {\\pi \\over 4})^2]\\)\\(=> 2[1 -{1\\over 2}\\times {1\\over 2}]\\)\\(=> 2[1 -{1\\over 4}] => 2 \\times {3\\over 4} = {3\\over 2} = RHS \\)Hence Proved | |