The value of sin2-4-sin23-4-sin25-4-sin27-4 is

Given nbsp; sin^2π4+sin^23π4+sin^25π4+sin^27π4 =sin^2π4+sin^2(π π4)+sin^2(π+π4)+sin^2(2 π π4) =sin^2π4+sin^2π4+( sinπ4)^2+( sinπ4)^2 =4sin^2π4=2 ...

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The value of ${sin}^{2} \frac{π}{4} + {sin}^{2} \frac{3 π}{4} + {sin}^{2} \frac{5 π}{4} + {sin}^{2} \frac{7 π}{4}$ is

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\frac{1}{2}

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{sin}^{2} \frac{π}{4} + {sin}^{2} \frac{3 π}{4} + {sin}^{2} \frac{5 π}{4} + {sin}^{2} \frac{7 π}{4}

s i n^{2} (\frac{π}{4}) + s i n^{2} (\frac{3 π}{4}) + s i n^{2} (\frac{5 π}{4}) + s i n^{2} (\frac{7 π}{4})

{sin}^{2} \frac{π}{4} + {sin}^{2} \frac{3 π}{4} + {sin}^{2} \frac{5 π}{4} + {sin}^{2} \frac{7 π}{4}

{sin}^{2} \frac{π}{8} + {sin}^{2} \frac{3 π}{8} + {sin}^{2} \frac{5 π}{8} + {sin}^{2} \frac{7 π}{8}

Prove that:

$s i n^{2} \frac{π}{8} + s i n^{2} \frac{3 π}{8} + s i n^{2} \frac{5 π}{8} + s i n^{2} \frac{7 π}{8} = 2$

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