If k -sin-18-sin5 -18-sin7 -18 then the numerical value of k isA- 1-4B- 1-8C- 1-16D- None of these

The correct option is B We have k=sin π/18sin 5 π/18 sin 7 π/18 = cos ( π/2 π/18 ) cos ( π/2 5 π/18 ) cos ( π/2 7 π/18 ) =cosπ/9 cos 2 π/9cos 4 π/9=sin 2 ...

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Byju's Answer

Standard XI

Mathematics

Product of Trigonometric Ratios in Terms of Their Sum

If k =sinπ/18...

Question

If $k = s i n \frac{π}{18} . s i n \frac{5 π}{18} . s i n \frac{7 π}{18}$ then the numerical value of k is

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Solution

The correct option is B

$W e h a v e k = s i n \frac{π}{18} s i n \frac{5 π}{18} s i n \frac{7 π}{18} = c o s (\frac{π}{2} - \frac{π}{18}) c o s (\frac{π}{2} - \frac{5 π}{18}) c o s (\frac{π}{2} - \frac{7 π}{18}) = c o s \frac{π}{9} c o s \frac{2 π}{9} c o s \frac{4 π}{9} = \frac{s i n 2^{3} \frac{π}{9}}{2^{3} s i n \frac{π}{9}} = \frac{s i n \frac{8 π}{9}}{8 s i n \frac{π}{9}} \frac{s i n (π - \frac{π}{9})}{8 s i n \frac{π}{9}} = \frac{1}{8}$

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If k=sinπ18.sin5π18.sin7π18 then the numerical value of k is

The correct option is B We have k=sin π18sin 5π18 sin 7π18=cos (π2−π18) cos (π2−5π18) cos (π2−7π18)=cosπ9 cos 2π9cos4π9=sin23 π923 sin π9=sin8π98 sin π9sin(π−π9)8 sin π9=18

If $k = s i n \frac{π}{18} . s i n \frac{5 π}{18} . s i n \frac{7 π}{18}$ then the numerical value of k is