The value of \(\displaystyle{{\cos}^{{4}}{\frac{{\pi}}{{{8}}}}}+{{\cos}^{{4}}{\frac{{{3}\pi}}{{{8}}}}}+{{\cos}^{{4}}{\frac{{{5}\pi}}{{{8}}}}}+{{\cos}^{{4}}{\frac{{{7}\pi}}{{{8}}}}}\)

Mary Bates

Mary Bates

Answered question

2022-04-10

The value of cos4π8+cos43π8+cos45π8+cos47π8

Answer & Explanation

embemiaEffoset4rs

embemiaEffoset4rs

Beginner2022-04-11Added 14 answers

cos4x=2cos22x1=2(2cos2x1)21=8cos4x8cos2x+1
If cos4x=0,4x=(2n+1)π2 where n is any integer
So the roots of 8c48c2+1=0 are cos(2n+1)π8 where n=0,1,2,3
If we set c4=d, we have 8d+1=8c2(8d+1)2=(8c2)2=64d
64d248d+1=0 (1)
As cosrπ8=cos(πrπ8)=cos(8r)π8,cos2rπ8
=cos2(8r)π8
n=03cos4(2n+1)π8=2n=01cos4(2n+1)π8
Using Vieta's formula on (1) to get
n=01cos4(2n+1)π8=34

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