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Question

cos4π8+cos43π8+cos45π8+cos47π8=32

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Solution

cos4π8+cos43π8+cos45π8+cos47π8
=cos4π8+cos4(π2π8)+cos4(π2+π8)+cos4(ππ8)
=cos4π8+{cos(π2π8)}4+{cos(π2+π8)}4+{cos(ππ8)}4
=cos4π8+sin4π8+(sinπ8)4+(cosπ8)4
=cos4π8+sin4π8+sin4π8+cos4π8
=2cos4π8+2sin4π8
=2(cos2π8)2+2(sin2π8)2
=2(1+cosπ82)2+2(1cos2.π82)2 [cos2x=12sin2x,cos2x=2cos2x1]
=2(1+cosπ42)2+2(1cosπ42)2
=24(1+12)2+24(112)2
=24{(1+12)2+(112)2}
=12{1+(12)2+22+1+(12)222}
=12{2+212}
=12×3=32

1199639_1339453_ans_67f1b72bb74a47bcbf7cc48fdbd8217e.JPG

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