# Could someone help me solve this trigonometry problem?

If $\theta \in \mathbb{R}$ such that $$\frac { 2\sec { \theta } +3\tan { \theta } +5\sin { \theta } -7\cos { \theta } +5 }{ 2\tan { \theta } +3\sec { \theta } +5\cos { \theta } +7\sin { \theta } +8 } =2\cos { \theta }$$ then $\sin { \theta } =?$ No idea how to solve this. If someone could show me the first one or two steps to push me in the right direction that would be great.

• Are you looking for a numerical value for $\sin\theta$. – John_dydx Jul 26 '15 at 13:55
• yes find sinθ. I have not idea to find it. – Tar'Tar Greatar Jul 26 '15 at 14:00
• Maybe the value of sinθ in term of Trigonometric function. – Tar'Tar Greatar Jul 26 '15 at 14:06
• Not too sure if this works, but try expressing everything in terms of $sine$ and $cosine$. Might make things easier to see. – Cataline Jul 26 '15 at 14:12
• I'm not sure you can find a numeric value for $\sin\theta$, the equation just looks very complicated. – John_dydx Jul 26 '15 at 14:28

subtracting $2\cos(x)$ on both sides and find a common denominator we get $$-{\frac {6\,\cos \left( x \right) \sec \left( x \right) +4\,\cos \left( x \right) \tan \left( x \right) +14\,\cos \left( x \right) \sin \left( x \right) +10\, \left( \cos \left( x \right) \right) ^{2} -2\,\sec \left( x \right) -3\,\tan \left( x \right) -5\,\sin \left( x \right) +23\,\cos \left( x \right) -5}{2\,\tan \left( x \right) +3\, \sec \left( x \right) +5\,\cos \left( x \right) +7\,\sin \left( x \right) +8}} =0$$ the simplified denominator is given by $$-14\, \left( \cos \left( x \right) \right) ^{2}\sin \left( x \right) -10\, \left( \cos \left( x \right) \right) ^{3}+\cos \left( x \right) \sin \left( x \right) -23\, \left( \cos \left( x \right) \right) ^{2}+3\,\sin \left( x \right) -\cos \left( x \right) +2 =0$$ solving for $\sin(x)$ we obtain $$\sin(x)=-{\frac {10\, \left( \cos \left( x \right) \right) ^{3}+23\, \left( \cos \left( x \right) \right) ^{2}+\cos \left( x \right) -2}{-3+14\, \left( \cos \left( x \right) \right) ^{2}-\cos \left( x \right) }}$$