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The value of \cos \frac{\pi}{19}+\cos \frac{3\pi}{19}+\cos \frac{5\pi}{19}+.....+\cos\frac{17\pi}{19} is equal to  (up to one decimal point):

 

Option: 1

0.5


Option: 2

0


Option: 3

1


Option: 4

0.7


Answers (1)

best_answer

 

Allied Angles -

- wherein

The trigonometric ratios for angles in all the four quadrants.

 

 

\cos \frac{\pi}{19}+\cos \frac{3\pi}{19}+\cos \frac{5\pi}{19}+.....+\cos \frac{17\pi}{19}

Here, A=\frac{\pi}{19},D=\frac{2\pi}{19},n=9

\because \cos A+\cos (A+D)+\cos (A+2D)+.........+\cos (A+(n-1)D)

=\frac{\sin\left ( \frac{nD}{2} \right ) }{\sin\frac{D}{2}}.\cos \left ( \frac{2A+(n-1)D}{2} \right )=\frac{\sin9\times \frac{\pi}{19}}{\sin\frac{\pi}{19}}\times \cos \left ( \frac{\frac{\pi}{19}+\frac{17\pi}{19}}{2} \right )

=\frac{\sin \frac{9\pi}{19}}{\sin\frac{\pi}{19}}\times \cos \frac{9\pi}{19}

=\frac{1}{2}.\frac{\sin \left ( \frac{18\pi}{19} \right )}{\sin \frac{\pi}{19}}=\frac{1}{2}.\frac{\sin \frac{\pi}{19}}{\sin \frac{\pi}{19}}=\frac{1}{2}

Posted by

manish painkra

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