# TRANSFORMATIONS OF SUMS AND PRODUCTS (Part-1)

### Prove the following identities :

$\,1(i).\,\,\,\sin10^{\circ}\sin 50^{\circ}+\sin50^{\circ}\sin250^{\circ}\\+\sin250^{\circ}\sin10^{\circ}=-\frac34$

Sol.  $\,\,\,\sin10^{\circ}\sin 50^{\circ}+\sin50^{\circ}\sin250^{\circ}\\+\sin250^{\circ}\sin10^{\circ}\\=\frac 12(2\sin10^{\circ}\sin50^{\circ})+\frac 12(2\sin50^{\circ}\sin 250^{\circ})\\+\frac 12(2\sin250^{\circ}\sin10^{\circ})\\=\frac 12\left[\cos(50^{\circ}-10^{\circ})-\cos(50^{\circ}+10^{\circ})\\+\cos(250^{\circ}-50^{\circ})-\cos(250^{\circ}+50^{\circ})\\+\cos(250^{\circ}-10^{\circ})-\cos(250^{\circ}+10^{\circ})\right]\\=\frac 12[\cos40^{\circ}-\cos60^{\circ}+\cos200^{\circ}-\cos300^{\circ}\\+\cos240^{\circ}-\cos260^{\circ}]\\=\frac 12\left[\cos40^{\circ}-\frac 12+\cos(2\times90^{\circ}+20^{\circ})\\-\cos(4\times 90^{\circ}-60^{\circ})+\cos(3 \times 90^{\circ}-30^{\circ})\\-\cos(3\times 90^{\circ}-10^{\circ})\right]\\=\frac12 \left[\cos40^{\circ}-\frac 12-\cos20^{\circ}-\cos60^{\circ} \\-\sin30^{\circ}+\sin10^{\circ}\right]\\=\frac12\left(\cos40^{\circ}-\frac 12-\cos20^{\circ}-\frac 12-\frac 12\\+\cos(90^{\circ}-10^{\circ})\right)\\=\frac 12\left(\cos40^{\circ}+\cos80^{\circ}-\cos20^{\circ}-\frac 32\right)\\=\frac 12\left(\sin(90^{\circ}-40^{\circ})+2\sin\frac{80^{\circ}+20^{\circ}}{2}\sin\frac{20^{\circ}-80^{\circ}}{2}\\-\frac 32\right)\\=\frac 12\left(\sin50^{\circ}+2\sin50^{\circ}\sin(-30^{\circ})-\frac 32\right)\\=\frac 12\left(\sin 50^{\circ}+2\sin50^{\circ}(-\frac 12)-\frac 32\right)\\=\frac 12\left(\sin50^{\circ}-\sin50^{\circ}-\frac 32\right)\\=-\frac 34\,\,\text{(proved)}$

$\,1(ii)\,\,\frac{\sin\alpha\sin11\alpha+\sin3\alpha\sin7\alpha}{\sin\alpha\cos11\alpha+\sin3\alpha\cos7\alpha}=\tan8\alpha$

Sol. $\,\,\frac{\sin\alpha\sin11\alpha+\sin3\alpha\sin7\alpha}{\sin\alpha\cos11\alpha+\sin3\alpha\cos7\alpha}\\=\frac{(2\sin\alpha\sin11\alpha)+(2\sin3\alpha\sin7\alpha)}{(2\sin\alpha\cos11\alpha)+(2\sin3\alpha\cos7\alpha)}\\=\frac{\cos(11\alpha-\alpha)-\cos(11\alpha+\alpha)+\cos(7\alpha-3\alpha)-\cos(7\alpha+3\alpha)}{\sin(11\alpha+\alpha)-\sin(11\alpha-\alpha)+\sin(7\alpha+3\alpha)-\sin(7\alpha-3\alpha)}\\=\frac{\cos10\alpha-\cos12\alpha+\cos4\alpha-\cos10\alpha}{\sin12\alpha-\sin10\alpha+\sin10\alpha-\sin4\alpha}\\=\frac{\cos4\alpha-\cos12\alpha}{\sin12\alpha-\sin4\alpha}\\=\frac{2\sin\frac{12\alpha+4\alpha}{2}\sin\frac{12\alpha-4\alpha}{2}}{2\sin\frac{12\alpha-4\alpha}{2}\cos\frac{12\alpha+4\alpha}{2}}\\=\tan8\alpha$

$\,1(iii)\,\,\,\tan\theta\tan(\frac{\pi}{3}+\theta)\tan(\frac{\pi}{3}-\theta)\\=\tan\theta \times \frac{\sin(\frac{\pi}{3}+\theta)\sin(\frac{\pi}{3}-\theta)}{\cos(\frac{\pi}{3}+\theta)\cos(\frac{\pi}{3}-\theta)}\\=\tan\theta \times \frac{2\sin(\frac{\pi}{3}+\theta)\sin(\frac{\pi}{3}-\theta)}{2\cos(\frac{\pi}{3}+\theta)\cos(\frac{\pi}{3}-\theta)}\\=\tan\theta \times \frac{\cos2\theta-\cos120^{\circ}}{\cos120^{\circ}+\cos2\theta}\\=\tan\theta\times \frac{\cos2\theta+\frac 12}{\cos2\theta-\frac 12}\\=\tan\theta \times \frac{2\cos2\theta+1}{2\cos2\theta-1}\\=\frac{\sin\theta}{\cos\theta} \times \frac{2\cos2\theta+1}{2\cos2\theta-1}\\=\frac{2\sin\theta\cos2\theta+\sin\theta}{2\cos\theta\cos2\theta-\cos\theta}\\=\frac{\sin3\theta-\sin\theta+\sin\theta}{\cos3\theta+\cos\theta-\cos\theta}\\=\tan3\theta$

$\,2(i)\,$ If $\,\,a\cos\phi=b\cos\theta,\,\,$ show that , $\,\,a\tan\theta+b\tan\phi=(a+b)\tan\frac{\theta+\phi}{2}.$

Sol.  $\,\,a\tan\theta+b\tan\phi\\=\frac{\left(\frac{b\cos\theta}{\cos\phi}\right)\frac{\sin\theta}{\cos\theta}+b\frac{\sin\phi}{\cos\phi}}{\left(\frac{b\cos\theta}{\cos\phi}\right)+b}\\=\frac{b\sin\theta+b\sin\phi}{b\cos\theta+b\cos\phi}\\=\frac{2\sin\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}}{2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}}\\=\tan\frac{\theta+\phi}{2} \\ \Rightarrow a\tan\theta+b\tan\phi=(a+b)\tan\frac{\theta+\phi}{2}$

$\,2(ii)\,$ If $\,\,p\sin\alpha=q\sin(120^{\circ}+\alpha)=r\sin(240^{\circ}+\alpha),\,\,$ prove that , $\,\,pq+qr+rp=0.$

Sol.  Let$\,\,p\sin\alpha=q\sin(120^{\circ}+\alpha)\\=r\sin(240^{\circ}+\alpha)=k(\neq 0) \\ \therefore \frac kp+\frac kq+\frac kr\\=\sin\alpha+\sin(120^{\circ}+\alpha)+\sin(240^{\circ}+\alpha)\\=\sin\alpha+2\sin(180^{\circ}+\alpha)\cos60^{\circ}\,\,[\text{By (*)}]\\=\sin\alpha-2\sin\alpha.\frac 12\\=\sin\alpha-\sin\alpha\\=0$

Note[*]: $\,\,\sin(120^{\circ}+\alpha)+\sin(240^{\circ}+\alpha)\\=2\sin\frac{(120^{\circ}+\alpha)+(240^{\circ}+\alpha)}{2}\cos\frac{(240^{\circ}+\alpha)-(120^{\circ}+\alpha)}{2}\\=2\sin(180^{\circ}+\alpha)\cos60^{\circ}$

$\,2(iii)\,$ If $\,\,\cos\theta=n\cos(\theta+2\phi),\,$ show that , $\,(n+1)\tan(\theta+\phi)=(n-1)\cot\phi$

Sol. We have ,$\,\,\cos\theta=n\cos(\theta+2\phi) \cdots(1)$

Now, $\,\,\frac{n+1}{n-1}\\=\frac{\frac{\cos\theta}{\cos(\theta+2\phi)}+1}{\frac{\cos\theta}{\cos(\theta+2\phi)}-1},\,\,[\text{By (1)}]\\=\frac{\cos\theta+\cos(\theta+2\phi)}{\cos\theta-\cos(\theta+2\phi)}\\=\frac{2\cos(\theta+\phi)\cos\phi}{2\sin(\theta+\phi)\sin\phi}\\=\cot(\theta+\phi)\cot\phi \\ \therefore (n+1)\tan(\theta+\phi)=(n-1)\cot\phi.$

$\,3.\,$ Prove that, $\,\,\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)\\=4\sin\frac{\alpha+\beta}{2}\sin\frac{\beta+\gamma}{2}\sin\frac{\gamma+\alpha}{2}$

Sol. $\,\,\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)\\=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+2\cos\frac{\alpha+\beta+2\gamma}{2}\sin\left(-\frac{\alpha+\beta}{2}\right)\\=2\sin\frac{\alpha=\beta}{2} \left(\cos\frac{\alpha-\beta}{2}-\cos\frac{\alpha+\beta+2\gamma}{2}\right)\\=4\sin\frac{\alpha+\beta}{2}\sin\frac{\beta+\gamma}{2}\sin\frac{\gamma+\alpha}{2}\,\, [**]$

Note[**] : $\,\,\cos C-\cos D=2\sin\frac{C+D}{2}\sin\frac{D-C}{2}$

$\,4.\,$ Express $\,\,4\sin A \cos B\cos C\,\,$ as the sum of four sines.

Sol. $\,\,4\sin A \cos B\cos C \\=2(2\sin A \cos B)\cos C\\=2\left[\sin(A+B)+\sin(A-B)\right]\cos C\\=2\sin(A+B)\cos C+2\sin(A-B)\cos C\\=\sin(A+B+C)+\sin(A+B-C)\\+\sin(A-B+C)+\sin(A-B-C)$

$\,5.\,$ Express $\,\,\cos\alpha+\cos\beta+\cos\gamma+\cos(\alpha+\beta+\gamma)\,\,$ as the product of three cosines.

Sol. $\,\,\cos\alpha+\cos\beta+\cos\gamma+\cos(\alpha+\beta+\gamma)\\=(\cos\alpha+\cos\beta)+\left[\cos\gamma+\cos(\alpha+\beta+\gamma)\right]\\=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+2\cos\frac{\alpha+\beta+2\gamma}{2}\cos\frac{\alpha+\beta}{2}\\=2\cos\frac{\alpha+\beta}{2}\left[\cos\frac{\alpha-\beta}{2}+\cos\frac{\alpha+\beta+2\gamma}{2}\right]\\=2\cos\frac{\alpha+\beta}{2}.2\cos\frac{\alpha+\gamma}{2}\cos\frac{\beta+\gamma}{2}\\=4\cos\frac{\alpha+\beta}{2}\cos\frac{\beta+\gamma}{2}\cos\frac{\gamma+\alpha}{2}$

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