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Question Number 134581 by EDWIN88 last updated on 05/Mar/21

$$\mathrm{Given}\{\begin{array}{l}\sin \mathrm{A}+\sin \mathrm{B}=0.7\\ \cos \mathrm{A}+\cos \mathrm{B}=0.8\end{array}$$$$\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{A}\mathrm{and}\mathrm{B}.$$

Answered by liberty last updated on 05/Mar/21

$$\iff 2+2\cos (\mathrm{A}-\mathrm{B})=\frac{49+64}{100}$$$$\iff 2\cos (\mathrm{A}-\mathrm{B})=-\frac{87}{100}$$$$\iff \cos (\mathrm{A}-\mathrm{B})=-\frac{87}{200}$$$$\iff \mathrm{A}-\mathrm{B}\approx 115.8°+\mathrm{k}.360°$$$$\iff \mathrm{A}\approx \mathrm{B}+115.8°+\mathrm{k}.360°,\mathrm{k}ϵ\mathbb{Z}$$$$$$

Answered by mr W last updated on 05/Mar/21

$$\sin A+\sin B=p(0.7)$$$$\cos A+\cos B=q(0.8)$$$$A=\frac{A+B}{2}+\frac{A-B}{2}=u+v$$$$B=\frac{A+B}{2}-\frac{A-B}{2}=u-v$$$$\sin (u+v)+\sin (u-v)=p$$$$\Rightarrow \sin u\cos v=\frac{p}{2}...\left(I\right)$$$$\cos (u+v)+\cos (u-v)=q$$$$\Rightarrow \cos u\cos v=\frac{q}{2}...\left(II\right)$$$$\left(I\right)\mathbf{÷}\left(II\right):$$$$\tan u=\frac{p}{q}$$$$\Rightarrow u=m\pi +{\tan }^{-1}\frac{p}{q}$$$${\left(I\right)}^2+{\left(II\right)}^2:$$$${\cos }^{2}v=\frac{p^2+q^2}{4}$$$$\cos v=\pm \frac{\sqrt{p^2+q^2}}{2}$$$$\Rightarrow v=n\pi \pm {\cos }^{-1}\frac{\sqrt{p^2+q^2}}{2}$$$$\Rightarrow A=u+v=(m+n)\pi +{\tan }^{-1}\frac{p}{q}\pm {\cos }^{-1}\frac{\sqrt{p^2+q^2}}{2}$$$$\Rightarrow B=u-v=(m-n)\pi +{\tan }^{-1}\frac{p}{q}\mp {\cos }^{-1}\frac{\sqrt{p^2+q^2}}{2}$$

Answered by bobhans last updated on 05/Mar/21

$$\Rightarrow 2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{7}{10}$$$$\Rightarrow 2\cos \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{8}{10}$$$$\left(\mathrm{i}\right)\tan \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)=\frac{7}{8};\mathrm{A}+\mathrm{B}=82.24°+2\mathrm{k}\pi$$$$\left(\mathrm{ii}\right)2+2\cos (\mathrm{A}-\mathrm{B})=\frac{113}{100}$$$$\Rightarrow \cos (\mathrm{A}-\mathrm{B})=-\frac{87}{200};\mathrm{A}-\mathrm{B}=115.78°+2\mathrm{k}\pi$$$$\mathrm{Thus}\{\begin{array}{l}\mathrm{A}=99.01°+2\mathrm{k}\pi \\ \mathrm{B}=-16.77°+2\mathrm{k}\pi \end{array}$$$$$$

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