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Period of Trigonometric Ratios

The value of ...

Question

The value of
${cos}^{2} (\frac{π}{16}) + {cos}^{2} (\frac{3 π}{16}) + {cos}^{2} (\frac{5 π}{16}) + {cos}^{2} (\frac{7 π}{16})$ is

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Solution

From given expression we can see that
$\frac{π}{16}, \frac{7 π}{16} and \frac{3 π}{16}, \frac{5 π}{16}$ are complementry angles

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

$= [{cos}^{2} (\frac{π}{16}) + {cos}^{2} (\frac{7 π}{16})] + [{cos}^{2} (\frac{3 π}{16}) + {cos}^{2} (\frac{5 π}{16})]$

$= [{cos}^{2} (\frac{π}{16}) + {cos}^{2} (\frac{π}{2} - \frac{π}{16})] + [{cos}^{2} (\frac{3 π}{16}) + {cos}^{2} (\frac{π}{2} - \frac{3 π}{16})]$

$= [{cos}^{2} (\frac{π}{16}) + {sin}^{2} (\frac{π}{16})] + [{cos}^{2} (\frac{3 π}{16}) + {sin}^{2} (\frac{3 π}{16})]$
$= 1 + 1 = 2$

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Similar questions

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}

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

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

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

L = {sin}^{2} (\frac{π}{16}) - {sin}^{2} (\frac{π}{8})

M = {cos}^{2} (\frac{π}{16}) - {sin}^{2} (\frac{π}{8}),

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Period of Trigonometric Ratios

Standard XII Mathematics

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