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Question Number 102390 by Dwaipayan Shikari last updated on 08/Jul/20

$$asin\theta +bcos\theta =c$$$$acosec\theta +bsec\theta =c$$$$$$$$provethatsin2\theta =\frac{2ab}{c^2-a^2-b^2}$$

Commented by PRITHWISH SEN 2 last updated on 08/Jul/20

$$\mathbf{Just}\mathbf{multiply}\mathbf{the}\mathbf{two}\mathbf{equations}$$

Commented by bobhans last updated on 09/Jul/20

$$if{a}^2+b^2⩾c^{2}or{a}^2+b^2⩽c^2?$$

Commented by bemath last updated on 09/Jul/20

$$\left(2\right)\frac{a}{\sin \theta }+\frac{b}{\cos \theta }=c$$$$\frac{a\cos \theta +b\sin \theta }{\frac{1}{2}\sin 2\theta }=c\Rightarrow a\cos \theta +b\sin \theta =\frac{1}{2}c\sin 2\theta$$$$\left(1\right)\times \left(2\right)$$$$(a\sin \theta +b\cos \theta )(a\cos \theta +b\sin \theta )=\frac{1}{2}{c}^2\sin 2\theta$$$$\frac{1}{2}{a}^2\sin 2\theta +ab\sin {}^2\theta +ab\cos {}^2\theta +\frac{1}{2}{b}^2\sin 2\theta =\frac{1}{2}{c}^2\sin 2\theta$$$$\{\frac{1}{2}{a}^2+\frac{1}{2}{b}^2\}\sin 2\theta +ab=\frac{1}{2}{c}^2\sin 2\theta$$$$ab=\{\frac{1}{2}{c}^2-\frac{1}{2}{a}^2-\frac{1}{2}{b}^2\}\sin 2\theta$$$$\sin 2\theta =\frac{2ab}{c^2-a^2-b^2}$$

Commented by 1549442205 last updated on 09/Jul/20

Answered by Dwaipayan Shikari last updated on 08/Jul/20

$$(asin\theta +bcos\theta )(acos\theta +bsin\theta ).\frac{2}{sin2\theta }=c^2$$$$(a^{2}sin\theta cos\theta +ab+b^{2}sin\theta cos\theta )\frac{2}{sin2\theta }=c^2$$$$\frac{sin2\theta }{sin2\theta }(a^2+b^2)+\frac{2ab}{sin2\theta }=c^2$$$$sin2\theta =\frac{2ab}{c^2-a^2-b^2}\left(proved\right)$$

Commented by bobhans last updated on 09/Jul/20

$$If\{\begin{array}{l}2\sin \theta +\cos \theta =2\\ 2\csc \theta +\sec \theta =2\end{array}$$$$\iff \sin 2\theta =\frac{2.2.1}{4-4-1}=-4??$$$$itcorrect?$$

Commented by 1549442205 last updated on 09/Jul/20

$$\mathrm{but}\mathrm{the}\mathrm{your}\mathrm{system}\mathrm{has}\mathrm{no}\mathrm{roots}!$$$$\mathrm{the}\mathrm{first}\mathrm{eqs}\mathrm{implies}\mathrm{that}\sin 2\theta =\frac{24}{25}!$$

Commented by PRITHWISH SEN 2 last updated on 09/Jul/20

$$\mathrm{a},\mathrm{b},\mathrm{and}\mathrm{c}\mathrm{cannot}\mathrm{be}\mathrm{an}\mathrm{arbitary}\mathrm{choice}\mathrm{for}\mathrm{these}$$$$\mathrm{identities}.\mathrm{It}\mathrm{should}\mathrm{follow}\mathrm{the}\mathrm{certain}\mathrm{condition}$$$$\mathrm{that}$$$$-1⩽\frac{2\mathrm{ab}}{{\mathrm{c}}^2-{\mathrm{a}}^2-{\mathrm{b}}^2}⩽1$$$$\mathrm{such}\mathrm{pair}\mathrm{can}\mathrm{be}$$$$\mathrm{a}=1\mathrm{b}=2\mathrm{and}\mathrm{c}=5\mathrm{and}\mathrm{so}\mathrm{on}.$$

Commented by bemath last updated on 09/Jul/20

$$itdoesmeant{c}^2⩾a^2+b^2$$

Answered by 1549442205 last updated on 09/Jul/20

$$\left(1\right)\iff \frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\sin \theta +\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\cos \theta =\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$$$$\iff \sin (\phi +\theta )=\sin \left[{\sin }^{-1}\left(\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right)\right](\mathrm{where}\phi =\arccos \frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}})$$$$(\mathrm{with}\mathrm{the}\mathrm{condition}{\mathrm{a}}^2+{\mathrm{b}}^2⩾{\mathrm{c}}^2)$$$$\Rightarrow \phi +\theta ={\sin }^{-1}\left(\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right)+2\mathrm{k}\pi$$$$\Rightarrow \theta ={\sin }^{-1}\left(\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right)+2\mathrm{k}\pi -\phi$$$$\Rightarrow \sin \theta =\sin ({\sin }^{-1}\left(\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right)-\phi )$$$$=\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\cos \phi -\frac{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\sin \phi$$$$=\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}.\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}-\frac{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}.\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}$$$$=\frac{\mathrm{ac}-\mathrm{b}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}}{{\mathrm{a}}^2+{\mathrm{b}}^2}$$$$\Rightarrow \cos \theta =\frac{1}{{\mathrm{a}}^2+{\mathrm{b}}^2}\sqrt{{({\mathrm{a}}^2+{\mathrm{b}}^2)}^2-(\mathrm{ac}-\mathrm{b}\sqrt{{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2)}^2}}$$$$\Rightarrow \sin 2\theta =2\sin \theta \cos \theta =2\left(\frac{\mathrm{ac}-\mathrm{b}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}}{{\mathrm{a}}^2+{\mathrm{b}}^2}\right)\left(\frac{1}{{\mathrm{a}}^2+{\mathrm{b}}^2}\sqrt{{({\mathrm{a}}^2+{\mathrm{b}}^2)}^2-(\mathrm{ac}-\mathrm{b}\sqrt{{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2)}^2}}\right)=\mathrm{f}(\mathrm{a},\mathrm{b},\mathrm{c})$$$$$$$$$$

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