Answer to Question #157512 in Trigonometry for Samychou

Question

Given that sinu + sinq - (27/65) and tan[(u + q)/ 2] = (3/7), find cos[(u - q)/ 2]

Answer 1

Solution

$\qquad\qquad \begin{aligned} \sin u+\sin q&= \frac{27}{65}\\ 2\sin (\frac{u+q}{2}).\cos(\frac{u-q}{2})&= \small \frac{27}{65}........................(1)\\ \\ \sin(u+q)&=2\sin (\frac{u+q}{2}).\cos(\frac{u+q}{2}).........................(2)\\ \\ \sin(u+q)&= \frac{2\tan(\frac{u+q}{2})}{1-\tan^2(\frac{u+q}{2})}\\ &= \frac{2\times\frac{3}{7}}{1-\frac{9}{49}}\\ &= \frac{21}{20}.................................(3)\\ \\ (2)=(3)\\ \\ 2\sin(\frac{u+q}{2}).\cos(\frac{u+q}{2})&= \frac{21}{20}...............................(4)\\ \\ \tan(\frac{u+q}{2})&= \frac{3}{7}\\ \frac{\sin(\frac{u+q}{2})}{\cos(\frac{u+q}{2})}&= \frac{3}{7}\implies \cos(\frac{u+q}{2})=\frac{7}{3}\sin(\frac{u+q}{2})............(5)\\ \\ Substituting \,(5)\,in\,(4), \\ 2\sin(\frac{u+q}{2}).\frac{7}{3}.\sin(\frac{u+q}{2})&=\frac{21}{20}\\ \sin^2(\frac{u+q}{2})&=\frac{9}{40}\\ \sin(\frac{u+q}{2})&=\pm\frac{3}{2\sqrt{10}}\\ &=+0.474\,or\,-0.474\\ \\ According\,to\,(1),\\ When\,\,\,\sin(\frac{u+q}{2})=+0.474&\implies\cos(\frac{u-q}{2})=+0.438\\ When\,\,\,\sin(\frac{u+q}{2})=-0.474 &\implies\cos(\frac{u-q}{2})=-0.438 \end{aligned}$