Prove nbsp-that nbsp-cos nbsp- nbsp- pi- - 15 - cos 2 nbsp- pi- - 15 - cos 3 nbsp- pi- nbsp- 15 - cos 4 nbsp- pi- nbsp- 15 - cos nbsp-5 nbsp- pi- nbsp- 15 - cos 6 nbsp- pi- nbsp- 15 - cos 7 nbsp- pi- nbsp- 15 - nbsp-1- 2-7

LHS #160;cos #160; #960;15.cos2 #960;15.cos3 #960;15.cos4 #960;15.cos5 #960;15.cos6 #960;15.cos7 #960;15=cos #160; #960;15.cos2 #960;15. ...

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Standard XII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

Prove nbsp...

Question

Prove that cos π / 15 . cos 2 π / 15 . cos 3 π / 15 . cos 4 π / 15 . cos 5 π / 15 . cos 6 π / 15 . cos 7 π / 15 = 1/ 2^7

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Solution

$LHS \cos \frac{π}{15} . \cos \frac{2 π}{15} . \cos \frac{3 π}{15} . \cos \frac{4 π}{15} . \cos \frac{5 π}{15} . \cos \frac{6 π}{15} . \cos \frac{7 π}{15} = \cos \frac{π}{15} . \cos \frac{2 π}{15} . \cos \frac{3 π}{15} . \cos \frac{4 π}{15} . (\frac{1}{2}) . \cos \frac{6 π}{15} . \cos (π - \frac{8 π}{15}) = - \frac{1}{2} \cos \frac{π}{15} . \cos \frac{2 π}{15} . \cos \frac{3 π}{15} . \cos \frac{4 π}{15} . \cos \frac{6 π}{15} . \cos \frac{8 π}{15} (since \cos (π - x) = - \cos x) = - \frac{1}{2} (\cos \frac{π}{15} . \cos \frac{2 π}{15} . \cos \frac{4 π}{15} . \cos \frac{8 π}{15}) (\cos \frac{3 π}{15} . \cos \frac{6 π}{15}) = - \frac{1}{2} (\cos \frac{π}{15} . \cos \frac{2 π}{15} . \cos 2^{2} (\frac{π}{15}) . \cos 2^{3} (\frac{π}{15})) (\cos \frac{3 π}{15} . \cos 2 (\frac{3 π}{15})) = - \frac{1}{2} (\frac{\sin 2^{4} (\frac{π}{15})}{2^{4} \sin (\frac{π}{15})}) (\frac{\sin 2^{2} (\frac{3 π}{15})}{2^{2} \sin (\frac{3 π}{15})}) (since \cos A \cos 2 A \cos 2^{2} A \cos 2^{3} A . . . . . \cos 2^{n - 1} A = \frac{\sin 2^{n} A}{2^{n} \sin A}) = - \frac{1}{2} (\frac{\sin \frac{16 π}{15}}{16 \sin \frac{π}{15}}) (\frac{\sin \frac{12 π}{15}}{4 \sin \frac{3 π}{15}}) = - \frac{1}{2} (\frac{\sin (π + \frac{π}{15})}{16 \sin \frac{π}{15}}) (\frac{\sin (π - \frac{3 π}{15})}{4 \sin \frac{3 π}{15}}) = - \frac{1}{2} (\frac{- \sin \frac{π}{15}}{16 \sin \frac{π}{15}}) (\frac{\sin \frac{3 π}{15}}{4 \sin \frac{3 π}{15}}) = \frac{1}{128} = \frac{1}{2^{7}} = RHS Hence Proved .$

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Similar questions

Q. Prove that:

\cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{3 π}{15} \cos \frac{4 π}{15} \cos \frac{5 π}{15} \cos \frac{6 π}{15} \cos \frac{7 π}{15} = \frac{1}{128}

Prove that:

sin [\frac{π}{6} + A] . cos [\frac{π}{3} - B] + sin [\frac{π}{3} - B] . cos [\frac{π}{6} + A] = cos (A - B)

Prove that

cos \frac{π}{15} cos \frac{2 π}{15} cos \frac{3 π}{15} cos \frac{4 π}{15} cos \frac{5 π}{15} cos \frac{6 π}{15} cos \frac{7 π}{15} = \frac{1}{2^{7}}

Q. Prove that:

\cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{7 π}{15} = \frac{1}{16}

$c o s \frac{π}{15} c o s \frac{2 π}{15} c o s \frac{3 π}{15} c o s \frac{4 π}{15} c o s \frac{5 π}{15} c o s \frac{6 π}{15} c o s \frac{7 π}{15} = \frac{1}{128}$

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