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Question

Mathematics Question on Trigonometric Functions

The value of sinπ10sin13π10sin \frac{\pi}{10}sin \frac{13\pi}{10} is

A

12\frac{1}{2}

B

−12-\frac{1}{2}

C

−14-\frac{1}{4}

D

11

Answer

−14-\frac{1}{4}

Explanation

Solution

sinπ10sin13π10=sinπ10sin(π+3π10)sin \frac{\pi}{10} sin \frac{13\pi}{10}=sin \frac{\pi}{10} sin\left(\pi+\frac{3\pi}{10}\right) =−sinπ10sin3π10=-sin \frac{\pi}{10} sin \frac{3\pi}{10} =−12[2sinπ10sin3π10]=-\frac{1}{2}\left[2sin \frac{\pi}{10}sin \frac{3\pi}{10}\right] =−12[cos(π10−3π10)−cos(π10+3π10)]=-\frac{1}{2}\left[cos\left(\frac{\pi}{10}-\frac{3\pi}{10}\right)-cos\left(\frac{\pi}{10}+\frac{3\pi}{10}\right)\right] =−12[cos(−2π10)−cos(4π10)]=-\frac{1}{2}\left[cos\left(-\frac{2\pi}{10}\right)-cos\left(\frac{4\pi}{10}\right)\right] =−12[cosπ5−cos2π5]=-\frac{1}{2}\left[cos \frac{\pi}{5}-cos \frac{2\pi}{5}\right] =−12[cos36∘−cos72∘]=-\frac{1}{2}\left[cos36^{\circ}-cos72^{\circ}\right] =−12[5+14−5−14]=-\frac{1}{2}\left[\frac{\sqrt{5}+1}{4}-\frac{\sqrt{5}-1}{4}\right] =−12[24]=−14=-\frac{1}{2}\left[\frac{2}{4}\right]=-\frac{1}{4}