The value of sin 2-8-sin 23 -8-sin 25 -8-sin 27 -8

The correct option is D 2sin^2π8+sin^23π8+sin^25π8+sin^27π8 =(sin^2π8+sin^27π8)+(sin^23π8+sin^25π8) =(sin^2π8+sin^2 ( π π8))+(sin^23π8+sin^2(π 3π8)) =(sin^2 ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

The value of ...

Question

The value of ${sin}^{2} \frac{π}{8} + {sin}^{2} \frac{3 π}{8} + {sin}^{2} \frac{5 π}{8} + {sin}^{2} \frac{7 π}{8}$ is

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Solution

The correct option is C $2$
${sin}^{2} \frac{π}{8} + {sin}^{2} \frac{3 π}{8} + {sin}^{2} \frac{5 π}{8} + {sin}^{2} \frac{7 π}{8}$
$= ({sin}^{2} \frac{π}{8} + {sin}^{2} \frac{7 π}{8}) + ({sin}^{2} \frac{3 π}{8} + {sin}^{2} \frac{5 π}{8})$
$= ({sin}^{2} \frac{π}{8} + {sin}^{2} (π - \frac{π}{8})) + ({sin}^{2} \frac{3 π}{8} + {sin}^{2} (π - \frac{3 π}{8}))$
$= ({sin}^{2} \frac{π}{8} + {sin}^{2} \frac{π}{8}) + ({sin}^{2} \frac{3 π}{8} + {sin}^{2} \frac{3 π}{8}) = 2 ({sin}^{2} \frac{π}{8} + {sin}^{2} (\frac{π}{2} - \frac{π}{8})) = 2 ({sin}^{2} \frac{π}{8} + {cos}^{2} \frac{π}{8}) = 2$

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Similar questions

Q. Solve

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

Q. Prove sin2 π/8 + sin2 3π/8 + sin2 5π/8 + sin2 7π/8 = 2

Q. If

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

.Find

m

Prove that:

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

Q. The value of

{sin}^{2} \frac{π}{16} + {sin}^{2} \frac{3 π}{16} + {sin}^{2} \frac{5 π}{16} + {sin}^{2} \frac{7 π}{16}

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The value of sin2π8+sin23π8+sin25π8+sin27π8 is

The correct option is C 2sin2π8+sin23π8+sin25π8+sin27π8 =(sin2π8+sin27π8)+(sin23π8+sin25π8) =(sin2π8+sin2(π−π8))+(sin23π8+sin2(π−3π8)) =(sin2π8+sin2π8)+(sin23π8+sin23π8)=2(sin2π8+sin2(π2−π8))=2(sin2π8+cos2π8)=2

The value of ${sin}^{2} \frac{π}{8} + {sin}^{2} \frac{3 π}{8} + {sin}^{2} \frac{5 π}{8} + {sin}^{2} \frac{7 π}{8}$ is