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Question Number 185486 by sonukgindia last updated on 22/Jan/23

Answered by a.lgnaoui last updated on 23/Jan/23

$$\mathrm{A}-(\mathrm{B}+\mathrm{C})=0$$$$\mathrm{A}+\mathrm{B}+\mathrm{C}=2(\mathrm{B}+\mathrm{C})=2\mathrm{A}$$$$\mathrm{A}+\mathrm{B}-\mathrm{C}=\mathrm{A}-(\mathrm{B}+\mathrm{C})+2\mathrm{B}=2\mathrm{B}$$$$\mathrm{A}-\mathrm{B}+\mathrm{C}=\mathrm{A}-(\mathrm{B}+\mathrm{C})+2\mathrm{C}=2\mathrm{C}$$$$$$$$\frac{\sin (\mathrm{A}+\mathrm{B}+\mathrm{C})+\sin (\mathrm{A}+\mathrm{B}-\mathrm{C})+\sin (\mathrm{A}-\mathrm{B}+\mathrm{C})}{2\sin \mathrm{Acos}\mathrm{Bcos}\mathrm{C}}=$$$$\frac{\sin 2\mathrm{A}+\sin 2\mathrm{B}+\sin 2\mathrm{C}}{2\sin \mathrm{Acos}\mathrm{Bcos}\mathrm{C}}=$$$$\frac{2[\sin \mathrm{Acos}\mathrm{A}+\sin \mathrm{Bcos}\mathrm{B}+\sin \mathrm{Ccos}\mathrm{C}]}{2\sin \mathrm{Acos}\mathrm{Bcos}\mathrm{C}}=$$$$\frac{\cos \mathrm{A}}{\cos \mathrm{Bcos}\mathrm{C}}+\frac{\tan \mathrm{B}}{\sin \mathrm{Acos}\mathrm{C}}+\frac{\tan \mathrm{C}}{\sin \mathrm{Acos}\mathrm{B}}$$$$=\frac{\cos \mathrm{A}}{\cos \mathrm{Bcos}\mathrm{C}}+\frac{1}{\sin \mathrm{A}}(\frac{\tan \mathrm{B}}{\cos \mathrm{C}}+\frac{\tan \mathrm{C}}{\cos \mathrm{B}})$$$$=\frac{\cos \mathrm{A}}{\cos \mathrm{Bcos}\mathrm{C}}+\frac{1}{\sin \mathrm{A}}\left(\frac{\sin \mathrm{B}+\sin \mathrm{C}}{\cos \mathrm{Bcos}\mathrm{C}}\right)$$$$=\frac{1}{\cos \mathrm{Bcos}\mathrm{C}}[\cos \mathrm{A}+\frac{\sin \mathrm{B}+\sin \mathrm{C}}{\sin \mathrm{A}}]\left(1\right)$$$$\sin \mathrm{A}=\sin (\mathrm{B}+\mathrm{C})$$$$$$$$\left(1\right)=\frac{1}{\mathbf{cos}\mathbf{B}\mathbf{cos}\mathbf{C}}\left[(\mathbf{cos}(\mathbf{B}+\mathbf{C})+\frac{\mathbf{sin}\mathrm{B}+\mathbf{sin}\mathrm{C}}{\mathbf{sin}(\mathbf{B}+\mathbf{C})})\right]$$$$$$$$=\frac{1}{\mathbf{cos}\mathbf{A}+\mathbf{sinBsinC}}[\mathbf{cos}\mathbf{A}+\frac{\mathbf{sin}\mathbf{B}+\mathbf{sin}\mathbf{C}}{\mathbf{sin}\mathbf{A}}]$$

Commented by mr W last updated on 23/Jan/23

$$thisislikewhenyoutrytosimplify$$$$\frac{1}{2}to\frac{4}{2+\frac{3}{1-\frac{1}{2}}}.$$

Answered by Frix last updated on 23/Jan/23

$$a=b+c$$$$\sin (a+b+c)+\sin (a+b-c)+\sin (a-b+c)=$$$$=\sin (2b+2c)+\sin 2b+\sin 2c$$$$2\sin a\cos b\cos c=$$$$={2\cos }^{2}b\sin c\cos c+2\sin b\cos b{\cos }^{2}c=$$$$=\frac{\sin (2b+2c)+\sin 2b+\sin 2c}{2}$$$$\Rightarrow$$$$\mathrm{answer}\mathrm{is}2$$

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