Differentiation (Part-38) | S N De

$\,6(i)\,$ Prove that, $\,\,\cos\theta\cos(60^{\circ}-\theta)\cos(60^{\circ}+\theta)=\frac14\cos3\theta$

Sol.  $\,\,\cos\theta\cos(60^{\circ}-\theta)\cos(60^{\circ}+\theta)\\=\frac12\cos\theta[2\cos(60^{\circ}-\theta)\cos(60^{\circ}+\theta)]\\=\frac12\cos\theta[\cos(60^{\circ}+\theta+60^{\circ}-\theta)\\+\cos(60^{\circ}+\theta-60^{\circ}+\theta)]\\=\frac12\cos\theta[\cos120^{\circ}+\cos2\theta]\\=\frac12\cos\theta[\cos(90^{\circ}+30^{\circ})+\cos2\theta]\\=\frac12\cos\theta[-\sin30^{\circ}+\cos2\theta]\\=\frac 12\cos\theta\times \left(-\frac12\right)+\frac14\times 2\cos2\theta\cos\theta\\=-\frac14\cos\theta+\frac14\times [\cos(2\theta+\theta)+\cos(2\theta-\theta)]\\=-\frac14\cos\theta+\frac14 \times [\cos3\theta+\cos\theta]\\=-\frac 14\cos\theta+\frac14\cos3\theta+\frac14\cos\theta\\=\frac14\cos3\theta\,\,\text{(proved)}$

$\,6(ii)\,$ Prove that, $\,\,4\sin\theta\sin\left(\frac{\pi}{3}+\theta\right)\sin\left(\frac{\pi}{3}-\theta\right)=\sin3\theta$

Sol.  $\,\,4\sin\theta\sin\left(\frac{\pi}{3}+\theta\right)\sin\left(\frac{\pi}{3}-\theta\right)\\=2\sin\theta\times 2 \sin\left(\frac{\pi}{3}+\theta\right)\sin\left(\frac{\pi}{3}-\theta\right)\\=2\sin\theta \times \left[\cos\left(\frac{\pi}{3}+\theta-\frac{\pi}{3}+\theta\right)\\-\cos\left(\frac{\pi}{3}+\theta+\frac{\pi}{3}-\theta\right)\right]\\=2\sin\theta \times [\cos2\theta-\cos\left(\pi-\frac{\pi}{3}\right)]\\=2\sin\theta\cos2\theta-2\sin\theta\left(-\cos\frac{\pi}{3}\right)\\=\sin(2\theta+\theta)-\sin(2\theta-\theta)+2\sin\theta\times \frac 12\\=\sin3\theta-\sin\theta+\sin\theta\\=\sin3\theta$

$\,6(iii)\,$ Prove that, $\,\,\cos\alpha\cos(120^{\circ}+\alpha)\cos(240^{\circ}+\alpha)\\=\frac14\cos3\alpha$

Sol. $\,\,\cos\alpha\cos(120^{\circ}+\alpha)\cos(240^{\circ}+\alpha)\\=\frac 12 \cos\alpha\times 2\cos(120^{\circ}+\alpha)\cos(240^{\circ}+\alpha)\\=\frac12\cos\alpha\times [\cos(240^{\circ}+\alpha+120^{\circ}+\alpha)\\+\cos(240^{\circ}+\alpha-120^{\circ}-\alpha)]\\=\frac12\cos\alpha \times [\cos(360^{\circ}+2\alpha)+\cos120^{\circ}]\\=\frac12 \cos\alpha\times [\cos2\alpha+\cos(180^{\circ}-60^{\circ})]\\=\frac12\cos\alpha\times [\cos2\alpha-\cos60^{\circ}]\\=\frac14 \times 2\cos2\alpha\cos\alpha-\frac12\cos\alpha\times \frac 12\\=\frac 14[\cos(2\alpha+\alpha)+\cos(2\alpha-\alpha)]-\frac 14\cos\alpha\\=\frac 14[\cos3\alpha+\cos\alpha]-\frac 14\cos\alpha\\=\frac 14\cos3\alpha+\frac 14\cos\alpha-\frac 14\cos\alpha\\=\frac14\cos3\alpha\,\,\text{(proved)}$

$\,7\,$ Find the value : 

$(i)\,\,\sin78^{\circ}-\sin18^{\circ}+\sin30^{\circ}-\sin42^{\circ}\\=\sin78^{\circ}-\sin42^{\circ}-\sin18^{\circ}+\sin30^{\circ}\\=2\cos\frac{78^{\circ}+42^{\circ}}{2}\sin\frac{78^{\circ}-42^{\circ}}{2}-\sin18^{\circ}+\frac12\\=2\cos60^{\circ}\sin18^{\circ}-\sin18^{\circ}+\frac 12\\=2\times \frac12\sin18^{\circ}-\sin18^{\circ}+\frac 12\\=\frac12$

$(ii)\,\,\sec20^{\circ}\sec40^{\circ}\sec80^{\circ}\\=\frac{2}{(2\cos20^{\circ}\cos40^{\circ})\cos80^{\circ}}\\=\frac{2}{(\cos60+\cos20)\cos80}\,\,[**]\\=\frac{2}{(\frac 12+\cos20)\cos80}\\=\frac{2}{\frac12\cos80+\frac12\times 2\cos80\cos20}\\=\frac{4}{\cos80+(\cos100+\cos60)}\,\,[**]\\=\frac{4}{\sin(90-80)+\cos(90+10)+\frac12}\\=\frac{4}{\sin10-\sin10+\frac12}\\=8\,\,\,\text{(ans.)}$

Note[**] : $\,\,\,2\cos\alpha \cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta)$

$(iii)\,\,\sqrt3 \cot20^{\circ}\cot40^{\circ}\cot80^{\circ}\\=\sqrt3 \times \frac{\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}}{\sin20^{\circ}\sin40^{\circ}\sin80^{\circ}}\\=\sqrt3 \times \frac{(2\cos20^{\circ}\cos40^{\circ})\cos80^{\circ}}{(2\sin20^{\circ}\sin40^{\circ})\sin80^{\circ}}\\=\sqrt3\times \frac{(\cos60^{\circ}+\cos20^{\circ})\cos80^{\circ}}{(\cos20^{\circ}-\cos60^{\circ})\sin80^{\circ}}\\=\sqrt3\times \frac{\frac12\cos80^{\circ}+\frac12(2\cos80^{\circ}\cos20^{\circ})}{\frac12 \times (2\sin80^{\circ}\cos20^{\circ})-\frac12\sin80^{\circ}}\\=\sqrt3 \times \frac{\cos80^{\circ}+\cos(80^{\circ}+20^{\circ})+\cos(80^{\circ}-20^{\circ})}{\sin(80^{\circ}+20^{\circ})+\sin(80^{\circ}-20^{\circ})-\sin80^{\circ}}\\=\sqrt3 \times \frac{\cos80^{\circ}+\cos100^{\circ}+\cos60^{\circ}}{\sin100^{\circ}+\sin60^{\circ}-\sin80^{\circ}}\\=\sqrt3\times \frac{\sin(90^{\circ}-80^{\circ})+\cos(90^{\circ}+10^{\circ})+\cos60^{\circ}}{\sin(90^{\circ}+10^{\circ})+\sin60^{\circ}-\sin(90^{\circ}-10^{\circ})}\\=\sqrt3 \times \frac{\sin10^{\circ}-\sin10^{\circ}+\cos60^{\circ}}{\cos10^{\circ}+\sin60^{\circ}-\cos10^{\circ}}\\=\sqrt3 \times \frac{\cos60^{\circ}}{\sin60^{\circ}}\\=\sqrt3 \times \cot60^{\circ}\\=\sqrt3\times \frac{1}{\sqrt3}\\=1$

$(iv)\,\,\frac{1}{\sin10^{\circ}}-2\sin70^{\circ}\\=\frac{1-2\times 2\sin70^{\circ}\sin10^{\circ}}{2\sin10^{\circ}}\\=\frac{1-2\left[\cos(70^{\circ}-10^{\circ})-\cos(70^{\circ}+10^{\circ})\right]}{2\sin10^{\circ}}\\=\frac{1-2(\cos60^{\circ}-\cos80^{\circ})}{2\sin10^{\circ}}\\=\frac{1-2(\frac 12-\sin10^{\circ})}{2\sin10^{\circ}}\,\,[**]\\=\frac{1-1+2\sin10^{\circ}}{2\sin10^{\circ}}\\=1$

Note[**] : $\,\,\cos80^{\circ}=\sin(90^{\circ}-80^{\circ})=\sin10^{\circ}$

$\,8.\,$ Prove the following identities: 

$(i)\,\,4\sin23^{\circ}\sin37^{\circ}\sin83^{\circ}=\cos21^{\circ}$

Sol. $\,\,4\sin23^{\circ}\sin37^{\circ}\sin83^{\circ}\\=2\sin23^{\circ}(2\sin37^{\circ}\sin83^{\circ})\\=2\sin23^{\circ}[\cos(83^{\circ}-37^{\circ})-\cos(83^{\circ}+37^{\circ})]\\=2\sin23^{\circ} [\cos46^{\circ}-\cos120^{\circ}]\\=2\sin23^{\circ}\cos46^{\circ}-2\sin23^{\circ}\cos(90^{\circ}+30^{\circ})\\=2\cos46^{\circ}\sin23^{\circ}+2\sin23^{\circ}\sin30^{\circ}\\=\sin(46^{\circ}+23^{\circ})-\sin(46^{\circ}-23^{\circ})+2\sin23^{\circ}\times \frac 12\\=\sin69^{\circ}-\sin23^{\circ}+\sin23^{\circ}\\=\sin69^{\circ}\\=\cos(90^{\circ}-69^{\circ})\\=\cos21^{\circ}\,\,\text{(proved)}$

$(ii)\,\,\cos2A+\cos4A+\cos6A+\cos8A\\=4\cos A\cos2A\cos5A$

Sol. $\,\,\cos2A+\cos4A+\cos6A+\cos8A\\=(\cos2A+\cos6A)+(\cos4A+\cos8A)\\=2\cos\left(\frac{6A+2A}{2}\right)\cos\left(\frac{6A-2A}{2}\right)+2\cos\left(\frac{8A+4A}{2}\right)\cos \left(\frac{8A-4A}{2}\right)\\=2\cos4A\cos2A+2\cos6A\cos2A\\=2\cos2A(\cos4A+\cos6A)\\=2\cos2A\times 2\cos \left(\frac{6A+4A}{2}\right) \cos \left(\frac{6A-4A}{2}\right)\\=4\cos2A\cos5A\cos A\\=4\cos A\cos2A\cos5A\,\,\text{(proved)}$

$(iii)\,\,\frac{\sin(A+B)-2\sin A+\sin(A-B)}{\cos(A+B)-2\cos A+\cos(A-B)}=\tan A$

Sol. $\,\frac{\sin(A+B)-2\sin A+\sin(A-B)}{\cos(A+B)-2\cos A+\cos(A-B)}\\=\frac{[\sin(A+B)+\sin(A-B)]-2\sin A}{[\cos(A+B)+\cos(A-B)]-2\cos A}\\=\frac{2\sin A\cos B-2\sin A}{2\cos A\cos B-2\cos A}\\=\frac{2\sin A(\cos B-1)}{2\cos A(\cos B-1)}\\=\tan A\,\,\text{(proved)}$

$\,(iv)\,\,\frac{\sin\theta+\sin3\theta+\sin5\theta+\sin7\theta}{\cos\theta+\cos3\theta+\cos5\theta+\cos7\theta}=\tan4\theta$

Sol. $\,\frac{\sin\theta+\sin3\theta+\sin5\theta+\sin7\theta}{\cos\theta+\cos3\theta+\cos5\theta+\cos7\theta}\\=\frac{(\sin\theta+\sin5\theta)+(\sin3\theta+\sin7\theta)}{(\cos\theta+\cos5\theta)+(\cos3\theta+\cos7\theta)}\\=\frac{2\sin3\theta\cos2\theta+2\sin5\theta\cos2\theta}{2\cos3\theta\cos2\theta+2\cos5\theta\cos2\theta}\,\,[**]\\=\frac{2\cos2\theta(\sin3\theta+\sin5\theta)}{2\cos2\theta(\cos3\theta+\cos5\theta)}\\=\frac{2\sin4\theta\cos\theta}{2\cos4\theta\cos\theta}\\=\tan4\theta\,\,\text{(proved)}$

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

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