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Question Number 68414 by behi83417@gmail.com last updated on 10/Sep/19

$$\mathrm{Is}\mathrm{it}\mathrm{possible}\mathrm{to}\mathrm{find}\mathrm{any}\mathrm{value}\mathrm{for}$$$$\mathbf{a},\mathbf{b},\mathbf{c}\mathrm{from}\mathrm{below}\mathrm{system}\mathrm{of}\mathrm{equetions}?$$$$\{\begin{array}{l}\mathbf{sina}+\mathbf{sinb}=\mathbf{sinc}\\ \mathbf{cosa}+\mathbf{cosb}=\mathbf{cosc}\end{array}$$

Commented by kaivan.ahmadi last updated on 10/Sep/19

$$\{\begin{array}{l}sinacosa+sinbcosa=sinccosa\\ -sinacosa-sinacosb=-sinacosc\end{array}\Rightarrow$$$$$$$$sinbcosa-sinacosb=sinccosa-sinacosc\Rightarrow$$$$sin(b-a)=sin(c-a)\Rightarrow$$$$\{\begin{array}{l}b-a=2k\pi +(c-a)\\ b-a=(2k+1)\pi +(a-c)\end{array}$$$$$$$$\Rightarrow \{\begin{array}{l}b-c=2k\pi \\ b-2a+c=(2k+1)\pi \end{array}$$

Answered by Tanmay chaudhury last updated on 10/Sep/19

$$2sin\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=sinc$$$$2cos\left(\frac{a+b}{2}\right)cos\left(\frac{a-b}{2}\right)=cosc$$$$tan\left(\frac{a+b}{2}\right)=tanc$$$$\frac{a+b}{c}=n\pi +c$$$$a+b=2n\pi +2c....\left(1\right)$$$$4{sin}^2\left(\frac{a+b}{2}\right){cos}^2\left(\frac{a-b}{2}\right)={sin}^{2}c$$$$4{cos}^2\left(\frac{a+b}{2}\right){cos}^2\left(\frac{a-b}{2}\right)={cos}^{2}c$$$$addthem$$$$4{cos}^2\left(\frac{a-b}{2}\right)=1$$$$cos\left(\frac{a-b}{2}\right)=\pm \frac{1}{2}$$$$cosidering+sign$$$$cos\left(\frac{a-b}{2}\right)=cos\left(\frac{\pi }{3}\right)$$$$\frac{a-b}{2}=2n\pi +\frac{\pi }{3}$$$$cos\left(\frac{a-b}{2}\right)=-\frac{1}{2}=cos(\pi +\frac{\pi }{3})$$$$\left(\frac{\mathit{a}-\mathit{b}}{2}\right)=2\mathit{n}\pi +\pi +\frac{\pi }{3}$$$$orcos\left(\frac{a-b}{2}\right)=-\frac{1}{2}=cos(\pi -\frac{\pi }{3})$$$$\frac{a-b}{2}=2n\pi +\frac{2\pi }{3}$$$$\mathit{t}\mathit{h}\mathit{u}\mathit{s}\mathit{w}\mathit{e}\mathit{c}\mathit{a}\mathit{n}\mathit{f}\mathit{i}\mathit{n}\mathit{d}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{e}\mathit{o}\mathit{f}\mathit{a}\mathit{a}\mathit{n}\mathit{d}\mathit{b}\mathit{i}\mathit{n}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{s}$$$$\mathit{o}\mathit{f}\mathit{c}$$$$$$

Answered by mr W last updated on 10/Sep/19

$${\sin }^{2}a+{\sin }^{2}b+2\sin a\sin b={\sin }^{2}c$$$${\cos }^{2}a+{\cos }^{2}b+2\cos a\cos b={\cos }^{2}c$$$$2+2(\sin a\sin b+\cos a\cos b)=1$$$$\sin a\sin b+\cos a\cos b=-\frac{1}{2}$$$$\cos (a-b)=-\frac{1}{2}$$$$\Rightarrow a-b=(2n+1)\pi \pm \frac{\pi }{3}$$

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