Answer to Question #157512 in Trigonometry for Samychou

Question #157512
Given that sinu + sinq - (27/65) and tan[(u + q)/ 2] = (3/7), find cos[(u - q)/ 2]
1
Expert's answer
2021-02-16T07:56:37-0500

Solution


"\\qquad\\qquad\n\\begin{aligned}\n\\sin u+\\sin q&= \\frac{27}{65}\\\\\n2\\sin (\\frac{u+q}{2}).\\cos(\\frac{u-q}{2})&= \\small \\frac{27}{65}........................(1)\\\\\n\\\\\n\\sin(u+q)&=2\\sin (\\frac{u+q}{2}).\\cos(\\frac{u+q}{2}).........................(2)\\\\\n\\\\\n\\sin(u+q)&= \\frac{2\\tan(\\frac{u+q}{2})}{1-\\tan^2(\\frac{u+q}{2})}\\\\\n&= \\frac{2\\times\\frac{3}{7}}{1-\\frac{9}{49}}\\\\\n&= \\frac{21}{20}.................................(3)\\\\\n\\\\\n(2)=(3)\\\\\n\\\\\n2\\sin(\\frac{u+q}{2}).\\cos(\\frac{u+q}{2})&= \\frac{21}{20}...............................(4)\\\\\n\\\\\n\\tan(\\frac{u+q}{2})&= \\frac{3}{7}\\\\\n\\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)\\\\\n\\\\\nSubstituting \\,(5)\\,in\\,(4),\n\\\\\n2\\sin(\\frac{u+q}{2}).\\frac{7}{3}.\\sin(\\frac{u+q}{2})&=\\frac{21}{20}\\\\\n\\sin^2(\\frac{u+q}{2})&=\\frac{9}{40}\\\\\n\\sin(\\frac{u+q}{2})&=\\pm\\frac{3}{2\\sqrt{10}}\\\\\n&=+0.474\\,or\\,-0.474\\\\\n\\\\\nAccording\\,to\\,(1),\\\\\nWhen\\,\\,\\,\\sin(\\frac{u+q}{2})=+0.474&\\implies\\cos(\\frac{u-q}{2})=+0.438\\\\\nWhen\\,\\,\\,\\sin(\\frac{u+q}{2})=-0.474 &\\implies\\cos(\\frac{u-q}{2})=-0.438\n\\end{aligned}"




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