sin4π8+cos4π8+sin43π8+cos43π8+sin45π8+cos45π8+sin47π8+cos47π8=

sin4π8+cos4π8+sin43π8+cos43π8+sin45π8+cos45π8+sin47π8+cos47π8=4[sin4π8+cos4π8]=4[(sin2π8+cos2π8)2−2sin2π8cos2π8]=4−2[2sinπ8cosπ8]2=4−2sin2π4=4−2[12]2=4−1=3

sin4π8+cos4π8+sin43π8+cos43π8+sin45π8+cos45π8+sin47π8+cos47π8=

sin4π8+cos4π8+sin43π8+cos43π8+sin45π8+cos45π8+sin47π8+cos47π8=4[sin4π8+cos4π8]=4[(sin2π8+cos2π8)2−2sin2π8cos2π8]=4−2[2sinπ8cosπ8]2=4−2sin2π4=4−2[12]2=4−1=3

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$\sin^{4} \frac{π}{8} + \cos^{4} \frac{π}{8} + \sin^{4} \frac{3 π}{8} + \cos^{4} \frac{3 π}{8} + \sin^{4} \frac{5 π}{8} + \cos^{4} \frac{5 π}{8} + \sin^{4} \frac{7 π}{8} + \cos^{4} \frac{7 π}{8} =$

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$3$

$8$

$\frac{3}{2}$

$4$

answer is A.

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Detailed Solution

$\sin^{4} \frac{π}{8} + \cos^{4} \frac{π}{8} + \sin^{4} \frac{3 π}{8} + \cos^{4} \frac{3 π}{8} + \sin^{4} \frac{5 π}{8} + \cos^{4} \frac{5 π}{8} + \sin^{4} \frac{7 π}{8} + \cos^{4} \frac{7 π}{8}$
$= 4 [\sin^{4} \frac{π}{8} + \cos^{4} \frac{π}{8}]$
$= 4 [{(\sin^{2} \frac{π}{8} + \cos^{2} \frac{π}{8})}^{2} - 2 \sin^{2} \frac{π}{8} \cos^{2} \frac{π}{8}]$
$= 4 - 2 {[2 \sin \frac{π}{8} \cos \frac{π}{8}]}^{2}$
$= 4 - 2 \sin^{2} \frac{π}{4} = 4 - 2 {[\frac{1}{\sqrt{2}}]}^{2} = 4 - 1 = 3$