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Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

Prove that :i...

Question

Prove that :
(i) $c o s \frac{2 π}{7} c o s \frac{4 π}{7} c o s \frac{6 π}{7} = \frac{1}{8}$
(ii) $c o s \frac{π}{11} c o s \frac{2 π}{11} c o s \frac{3 π}{11} c o s \frac{4 π}{11} c o s \frac{5 π}{11} = \frac{1}{32}$

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Solution

i) $cos \frac{2 π}{7} cos \frac{4 π}{7} cos \frac{6 π}{7} = \frac{1}{8}$
$= \frac{1}{2 sin \frac{2 π}{7}} [2 sin (\frac{2 π}{7}) \cdot cos (\frac{2 π}{7})] cos (\frac{4 π}{7}) \cdot cos (\frac{6 π}{7})$
= $\frac{1}{2 sin \frac{2 π}{7}} [sin (\frac{4 π}{7}) \cdot cos (\frac{4 π}{7})] \cdot cos (\frac{6 π}{7})$
= $\frac{1}{4 sin \frac{2 π}{7}} [sin (\frac{8 π}{7}) \cdot cos (\frac{6 π}{7})]$

Using the formula of $[2 sin A cos B = sin (A + B) + sin (A - B)]$
= $\frac{1}{8 sin \frac{2 π}{7}} [2 sin (\frac{8 π}{7}) \cdot cos (\frac{6 π}{7})]$
= $\frac{1}{8 sin \frac{2 π}{7}} [sin (\frac{14 π}{7}) + sin (\frac{2 π}{7})]$

= $\frac{1}{8 sin \frac{2 π}{7}} [sin (2 π) + sin (\frac{2 π}{7})]$

= $\frac{1}{8}$ ......as $sin 2 π = 0$
ii) $c o s \frac{π}{11} c o s \frac{2 π}{11} c o s \frac{3 π}{11} c o s \frac{4 π}{11} c o s \frac{5 π}{11} = \frac{1}{32}$
= $\frac{2 sin \frac{π}{11} cos \frac{π}{11} c o s \frac{2 π}{11} c o s (π - \frac{8 π}{11}) c o s \frac{4 π}{11} c o s \frac{5 π}{11}}{2 sin \frac{π}{11}}$
= $\frac{- 2 sin \frac{2 π}{11} c o s \frac{2 π}{11} c o s (\frac{8 π}{11}) c o s \frac{4 π}{11} c o s \frac{5 π}{11}}{4 sin \frac{π}{11}}$
= $\frac{- 2 sin \frac{4 π}{11} c o s \frac{4 π}{11} c o s (\frac{8 π}{11}) c o s \frac{5 π}{11}}{8 sin \frac{π}{11}}$

= $\frac{- 2 sin \frac{8 π}{11} c o s (\frac{8 π}{11}) c o s \frac{5 π}{11}}{16 sin \frac{π}{11}}$
= $\frac{- sin \frac{16 π}{11} c o s \frac{5 π}{11}}{16 sin \frac{π}{11}}$

= $\frac{- sin (π + \frac{5 π}{11}) c o s \frac{5 π}{11}}{16 sin \frac{π}{11}}$

= $\frac{- sin (π + \frac{5 π}{11}) c o s \frac{5 π}{11}}{16 sin \frac{π}{11}}$

= $\frac{2 sin (\frac{5 π}{11}) c o s \frac{5 π}{11}}{32 sin \frac{π}{11}}$

= $\frac{sin (\frac{10 π}{11})}{32 sin \frac{π}{11}}$
= $\frac{sin (π - \frac{π}{11})}{32 sin \frac{π}{11}}$

= $\frac{1}{32}$

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Similar questions

cos \frac{π}{11} cos \frac{2 π}{11} cos \frac{3 π}{11} cos \frac{4 π}{11} cos \frac{5 π}{11} = \frac{1}{32}

c o s \frac{π}{11} c o s \frac{2 π}{11} c o s \frac{3 π}{11} c o s \frac{4 π}{11} c o s \frac{5 π}{11}

Q. Find the value of -

cos \frac{π}{11} cos \frac{2 π}{11} cos \frac{3 π}{11} cos \frac{4 π}{11} cos \frac{5 π}{11}

.

cos \frac{2 π}{7} + cos \frac{4 π}{7} + cos \frac{6 π}{7} = - \frac{1}{2}

cos \frac{2 π}{7} cos \frac{4 π}{7} cos \frac{6 π}{7} = \frac{1}{8}

. Then the numerical value of

{c o s e c}^{2} \frac{π}{7} + {c o s e c}^{2} \frac{2 π}{7} + {c o s e c}^{2} \frac{3 π}{7}

Q. The value of

cos \frac{2 π}{7} cos \frac{4 π}{7} cos \frac{6 π}{7}

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Tangent, Cotangent, Secant, Cosecant in Terms of Sine and Cosine

Standard XII Mathematics

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