Prove that-cos 2-8-cos 23 -8-cos 25 -8-cos 27 -8-2

LHS, cos^2 π/8+ cos^2 3 π/8 + cos^2 5 π/8+ cos^2 7 π/8 = cos^2 π/8+cos^2 3 π/8 + cos^2 (π 3 π/8 ) + cos^2 (π π/8 ) = cos^2 π/8+cos^2 3 π/8 + cos^2 3 π/8 + c ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

Prove that:co...

Question

Prove that:

$c o s^{2} \frac{π}{8} + c o s^{2} \frac{3 π}{8} + c o s^{2} \frac{5 π}{8} + c o s^{2} \frac{7 π}{8} = 2$

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Solution

LHS,

$c o s^{2} \frac{π}{8} + c o s^{2} \frac{3 π}{8} + c o s^{2} \frac{5 π}{8} + c o s^{2} \frac{7 π}{8} = c o s^{2} \frac{π}{8} + c o s^{2} \frac{3 π}{8} + c o s^{2} (π - \frac{3 π}{8}) + c o s^{2} (π - \frac{π}{8}) = c o s^{2} \frac{π}{8} + c o s^{2} \frac{3 π}{8} + c o s^{2} \frac{3 π}{8} + c o s^{2} \frac{π}{8} = 2 (c o s^{2} \frac{π}{8} + c o s^{2} \frac{3 π}{8}) = 2 (c o s^{2} \frac{π}{8} + c o s^{2} (\frac{π}{2} - \frac{π}{8})) = 2 (c o s^{2} \frac{π}{8} + s i n^{2} \frac{π}{8}) = 2 = RHS$

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Similar questions

Q. Prove the following identities:

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

Q. Solve

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

Q. Find the value of

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

Q. The value of

{cos}^{2} (\frac{π}{16}) + {cos}^{2} (\frac{3 π}{16}) + {cos}^{2} (\frac{5 π}{16}) + {cos}^{2} (\frac{7 π}{16})

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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Prove that: cos2 π8+cos23π8+cos25π8+cos27π8=2

LHS, cos2 π8+cos23π8+cos25π8+cos27π8=cos2π8+cos23π8+cos2(π−3π8)+cos2 (π−π8)=cos2 π8+cos2 3π8+cos2 3π8+cos2 π8=2(cos2 π8+cos2 3π8)=2(cos2 π8+cos2(π2−π8))=2(cos2 π8+sin2 π8)=2=RHS

Prove that:

$c o s^{2} \frac{π}{8} + c o s^{2} \frac{3 π}{8} + c o s^{2} \frac{5 π}{8} + c o s^{2} \frac{7 π}{8} = 2$