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# Solve: sin(x+11pi/6)-sin(x-11pi/6)=1

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I need understanding and help solving

Jun 29, 2018

#1
+8146
+2

$$\sin(x +\frac{11\pi}{6})\,-\,\sin(x -\frac{11\pi}{6})\,=\,1$$

We can use the sum of angles formula for sine:

$$\sin(\alpha \pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\\~\\ \ \\ {\color{BlueViolet} \sin(x +\frac{11\pi}{6})=\sin x\cos\frac{11\pi}{6}+\cos x\sin\frac{11\pi}{6}\quad=(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)}\\~\\ {\color{Fuchsia} \sin(x -\frac{11\pi}{6})=\sin x\cos\frac{11\pi}{6}-\cos x\sin\frac{11\pi}{6}\quad=(\sin x)(\frac{\sqrt3}{2})-(\cos x)(-\frac12)}$$

So if...

$${\color{BlueViolet}\sin(x +\frac{11\pi}{6})}\,-\,{\color{Fuchsia}\sin(x -\frac{11\pi}{6})}\,=\,1\\~\\ [\,{\color{BlueViolet}(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)}\,]\,-\,[\,{\color{Fuchsia}(\sin x)(\frac{\sqrt3}{2})-(\cos x)(-\frac12)}\,]\,=\,1\\~\\ (\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)-(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)\,=\,1\\~\\ 2(\cos x)(-\frac12)\,=\,1\\~\\ - \cos x\,=\,1\\~\\ \cos x\,=\,-1\\~\\ x\,=\,\pi+2\pi n\quad \text{, where }\ n\ \text{ is an integer.}$$      then...

Jun 29, 2018

#1
+8146
+2

$$\sin(x +\frac{11\pi}{6})\,-\,\sin(x -\frac{11\pi}{6})\,=\,1$$

We can use the sum of angles formula for sine:

$$\sin(\alpha \pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\\~\\ \ \\ {\color{BlueViolet} \sin(x +\frac{11\pi}{6})=\sin x\cos\frac{11\pi}{6}+\cos x\sin\frac{11\pi}{6}\quad=(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)}\\~\\ {\color{Fuchsia} \sin(x -\frac{11\pi}{6})=\sin x\cos\frac{11\pi}{6}-\cos x\sin\frac{11\pi}{6}\quad=(\sin x)(\frac{\sqrt3}{2})-(\cos x)(-\frac12)}$$

So if...

$${\color{BlueViolet}\sin(x +\frac{11\pi}{6})}\,-\,{\color{Fuchsia}\sin(x -\frac{11\pi}{6})}\,=\,1\\~\\ [\,{\color{BlueViolet}(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)}\,]\,-\,[\,{\color{Fuchsia}(\sin x)(\frac{\sqrt3}{2})-(\cos x)(-\frac12)}\,]\,=\,1\\~\\ (\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)-(\sin x)(\frac{\sqrt3}{2})+(\cos x)(-\frac12)\,=\,1\\~\\ 2(\cos x)(-\frac12)\,=\,1\\~\\ - \cos x\,=\,1\\~\\ \cos x\,=\,-1\\~\\ x\,=\,\pi+2\pi n\quad \text{, where }\ n\ \text{ is an integer.}$$      then...

hectictar Jun 29, 2018