Differentiation (Part-38) | S N De

 $\,1.\,$ For any $\,\triangle ABC,\,$ prove that, $\,(i)\,\,a\sin(A/2+B)=(b+c)\sin(A/2)$

Sol. We know, for any $\,\triangle ABC,\,\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Now, $\,\,\frac{a}{b+c}\\=\frac{2R\sin A}{2R\sin B+2R\sin C}\\=\frac{\sin A}{\sin B+\sin C}\\=\frac{2\sin(A/2)\cos(A/2)}{2\sin\left(\frac{B+C}{2}\right)\cos\left(\frac{C-B}{2}\right)}\\=\frac{\sin(A/2)\cos(A/2)}{\sin\left(\frac{\pi}{2}-\frac A2\right)\cos\left(\frac{\pi-A-B-B}{2}\right)}\\ ~~~~~~~~~~~~~~~[\because A+B+C=\pi]\\=\frac{\sin(A/2)\cos(A/2)}{\cos\frac{A}{2}\cos\left[\frac{\pi}{2}-\left(\frac A2+B\right)\right]}\\=\frac{\sin(A/2)}{\sin\left(\frac A2+B\right)} \\ \therefore \,\,a\sin(A/2+B)=(b+c)\sin(A/2)\,\,\text{(proved)}$

$\,(ii)\,\,(b-c)\cos\frac A2=a\sin\left(\frac{B-C}{2}\right)$

Sol. We know, for any $\,\triangle ABC,\,\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Now, $\,\,\frac{b-c}{a}\\=\frac{2R(\sin B-\sin C)}{2R\sin A}\\=\frac{2\sin\left(\frac{B-C}{2}\right)\cos\left(\frac{B+C}{2}\right)}{2\sin(A/2)\cos(A/2)}\\=\frac{\sin\left(\frac{B-C}{2}\right)\cos\left(\frac{\pi-A}{2}\right)}{\sin(A/2)\cos(A/2)}\\ ~~~~~~~~~~~~~~~[\because A+B+C=\pi]\\=\frac{\sin\left(\frac{B-C}{2}\right)\sin(A/2)}{\sin(A/2)\cos(A/2)}\\ \therefore (b-c)\cos(A/2)=a\sin\left(\frac{B-C}{2}\right)\,\,\,\text{(proved)}$

$\,(iii)\,\,a \cos A+b \cos B+c\cos C\\=4R\sin A\sin B\sin C$

Sol. We know, for any $\,\triangle ABC,\,\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Now, $\,\,a\cos A+b\cos B+c\cos C\\=2R\sin A\cos A+2R\sin B\cos B\\+2R\sin C\cos C\\=R(\sin2A+\sin2B)\\+2R\sin C\cos C\\=R. 2\sin(A+B)\cos( A-B)\\ +2R\sin C\cos C\\=2R.\sin(\pi-C)\cos(A-B)\\+2R\sin C\cos C\\=2R\sin C\left[\cos(A-B)+\cos C\right]\\=2R\sin C. 2\cos\left(\frac{A+C-B}{2}\right)\cos\left(\frac{A-B-C}{2}\right)\\=4R\sin C\cos\left(\frac{\pi-B-B}{2}\right)\cos\left(\frac{A-(\pi-A)}{2}\right)\\ ~~~~~~~~~~~~~~~~~~~~~~~~~[\because A+B+C=\pi]\\=4R\sin C\cos\left(\frac{\pi}{2}-B\right)\cos\left(\frac{\pi}{2}-A\right)\\=4R\sin A\sin B\sin C\,\,\,\text{(proved)}$

$\,(iv)\,\,a\sin(B-C)+b\sin(C-A) \\ +c\sin(A-B)=0$

Sol. We know , for any $\,\,\triangle ABC,\,\,\left[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\right]$

Now,$\,\,\,a\sin(B-C)+b\sin(C-A)+c\sin(A-B)\\=2R\sin A\sin(B-C)+2R\sin B\sin(C-A)\\+2R\sin C\sin(A-B)\\=2R[\sin A(\sin B\cos C-\cos B\sin C)\\+\sin B(\sin C\cos A-\cos C\sin A)\\ +\sin C(\sin A\cos B-\cos A\sin B)]\\=2R \times 0\\=0\,\,\,\text{(proved)}$

$\,(v).\,\,\frac{b^2-c^2}{\cos B+\cos C}+\frac{c^2-a^2}{\cos C+\cos A}+\frac{a^2-b^2}{\cos A+\cos B}=0$

Sol. We know , for any $\,\,\triangle ABC,\\ \left[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\right]$

Now, $\,\,\frac{b^2-c^2}{\cos B+\cos C}\\=\frac{4R^2(\sin^2B-\sin^2C)}{\cos B+\cos C}\\=4R^2.\frac{\cos^2C-\cos^2B}{\cos C+\cos B}\\=4R^2(\cos C-\cos B)$

Similarly, $\,\,\frac{c^2-a^2}{\cos C+\cos A}\\=4R^2(\cos A-\cos C)$

Finally, $\,\,\frac{a^2-b^2}{\cos A+\cos B}\\=4R^2(\cos B-\cos A) \\ \therefore \frac{b^2-c^2}{\cos B+\cos C}+\frac{c^2-a^2}{\cos C+\cos A}+\frac{a^2-b^2}{\cos A+\cos B} \\=4R^2 (\cos C-\cos B +\cos A-\cos C \\ + \cos B-\cos A)\\=4R^2 \times 0\\=0\,\,\text{(proved)}$

$\,(vi)\,\,a \cos B\cos C+b \cos C\cos A \\ +c \cos A \cos B=\frac{abc}{4R^2}$

Sol. We know, for any $\,\,\triangle ABC, \\  \left[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\right]$

$\,\,\therefore a\cos B\cos C +b \cos C\cos A \\ +c \cos A \cos B\\=\cos B(a \cos C+c \cos A)+b \cos C \cos A\\=\cos B. b+b \cos C\cos A \\ [\because b=a\cos C+c \cos A]\\=b\left[\cos \{\pi-(A+C)\}+\frac 12. 2\cos A\cos C\right]\\=b\left[-\cos(A+C)+\frac 12\{\cos(A+C)\\ +\cos(A-C)\}\right]\\=\frac b2\left[\cos(A-C)-\cos(A+C)\right]\\=\frac b2. 2\sin A\sin C\\=b.\frac{a}{2R}.\frac{c}{2R}\\=\frac{abc}{4R^2}\,\,\,\text{(proved)}$

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$\,(vii)\,\,\frac{b^2-c^2}{a^2}\sin2A+\frac{c^2-a^2}{b^2}\sin2B \\ +\frac{a^2-b^2}{c^2}\sin 2C  =0$

Sol. We know, for any $\,\,\triangle ABC,\\ \left[ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\right]$

Now, $\,\,\frac{b^2-c^2}{a^2}\sin2A\\=\frac{4R^2(\sin^2B-\sin^2C)}{4R^2 \sin^2A}.2\sin A\cos A\\=\sin(B+C)\sin(B-C).2\frac{\cos A}{\sin A}\\=\frac{2\sin(\pi-A)\sin(B-C)\cos \{\pi-(B+C)\}}{\sin A}\\=-2\sin(B-C)\cos(B+C)\\=2\sin(C-B)\cos(B+C)\\=\sin2C-\sin2B.$

Similarly, $\,\,\frac{c^2-a^2}{b^2}\sin2B=\sin2A-\sin2C$

Finally, $\,\,\frac{a^2-b^2}{c^2}\sin 2C=\sin2B-\sin2A$

$\therefore \,\,\frac{b^2-c^2}{a^2}\sin2A+\frac{c^2-a^2}{b^2}\sin2B+\frac{a^2-b^2}{c^2}\sin 2C\\=\sin2C-\sin2B+\sin2A\\-\sin2C+\sin2B-\sin2A\\=0\,\,\,\text{(proved)}$

$\,\,(viii)\,\,(b^2-c^2)\cos2A+(c^2-a^2)\cos2B \\ +(a^2-b^2)\cos2C=0$

Sol. We know, for any $\,\,\triangle ABC,\\  \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$\,\,(b^2-c^2)\cos2A\\=4R^2(\sin^2B-\sin^2C)\cos2A\\=2R^2(2\sin^2B-2\sin^2C)\cos2A\\=2R^2[(1-2\sin^2C)-(1-2\sin^2B)]\cos2A\\=2R^2(\cos2C-\cos2B)\cos2A\\=2R^2(\cos2C\cos2A-\cos2B\cos2A) \rightarrow(1)$

Similarly, $\,(c^2-a^2)\cos2B\\=2R^2(\cos2B\cos2A-\cos2B\cos2C)\rightarrow(2)$

Finally, $\,\, (a^2-b^2)\cos2C\\=2R^2(\cos2B\cos2C-\cos2A\cos2C)\rightarrow(3)$

Adding $\,(1),\,(2),\,(3)$ we get,

$\,\,(b^2-c^2)\cos2A+(c^2-a^2)\cos2B \\ +(a^2-b^2)\cos2C=0$

$\,(ix)\,\,\frac{a^2\sin(B-C)}{\sin A}+\frac{b^2\sin(C-A)}{\sin B}+\frac{c^2\sin(A-B)}{\sin C}=0$

Sol. We know, for any $\,\,\triangle ABC,\\  \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\rightarrow(1)$

Now, $\,\,\frac{a^2\sin(B-C)}{\sin A}\\=\frac{4R^2\sin^2A\sin(B-C)}{\sin A}\,\,[\text{By (1)}]\\=4R^2\sin A\sin(B-C)\\=4R^2\sin A(\sin B\cos C-\cos B\sin C)\rightarrow(2)$

Similarly, $\,\,\frac{b^2\sin(C-A)}{\sin B}\\=4R^2\sin B(\sin C\cos A-\cos C\sin A)\rightarrow(3)$

Finally, $\,\,\frac{c^2\sin(A-B)}{\sin C}\\=4R^2\sin C(\sin A\cos B-\cos A\sin B)\rightarrow(4)$

Adding $\,\,(2),\,(3),\,(4)\,\,$ we get,

$\,\frac{a^2\sin(B-C)}{\sin A}+\frac{b^2\sin(C-A)}{\sin B}+\frac{c^2\sin(A-B)}{\sin C}=0$