# TRANSFORMATIONS OF SUMS AND PRODUCTS (Part-4)

$\,3(i)\,$ Prove that $\,\,\sin19^{\circ}+\sin41^{\circ}+\sin83^{\circ}=\sin23^{\circ}+\sin37+\sin79^{\circ}$

Sol. $\,\,\text{L.H.S}=\sin19^{\circ}+\sin41^{\circ}+\sin83^{\circ}\\=2\sin\frac{19+41}{2}\cos\frac{41^{\circ}-19^{\circ}}{2}+\sin(90^{\circ}-7^{\circ})\\=2\sin30^{\circ}\cos11^{\circ}+\cos7^{\circ}\\=2.\frac12.\cos11^{\circ}+\cos7^{\circ}\\=\cos11^{\circ}+\cos7^{\circ}$

Again, $\,\text{R.H.S}=\,\sin23^{\circ}+\sin37^{\circ}+\sin79^{\circ}\\=2\sin\frac{23^{\circ}+37^{\circ}}{2}\cos\frac{37^{\circ}-23^{\circ}}{2}+\sin(90^{\circ}-11^{\circ})\\=2\sin30^{\circ}\cos7^{\circ}+\cos11^{\circ}\\=2.\frac 12.\cos7^{\circ}+\cos11^{\circ}\\=\cos7^{\circ}+\cos11^{\circ}$

Hence, L.H.S.=R.H.S

$\,3(ii)\,$ Prove that $\,\,\sin10^{\circ}+\sin20^{\circ}+\sin40^{\circ}+\sin50^{\circ}=\sin70^{\circ}+\sin80^{\circ}$

Sol. $\,\,\sin10^{\circ}+\sin20^{\circ}+\sin40^{\circ}+\sin50^{\circ}\\=(\sin10^{\circ}+\sin50^{\circ})+(\sin20^{\circ}+\sin40^{\circ})\\=2\sin\frac{50^{\circ}+10^{\circ}}{2}\cos\frac{50^{\circ}-10^{\circ}}{2}\\+2\sin\frac{40^{\circ}+20^{\circ}}{2}\cos\frac{40^{\circ}-20^{\circ}}{2}\\=2\sin30^{\circ}\cos20^{\circ}+2\sin30^{\circ}\cos10^{\circ}\\=2.\frac 12.\cos(90^{\circ}-70^{\circ})+2.\frac 12.\cos(90^{\circ}-80^{\circ})\\=\sin70^{\circ}+\sin80^{\circ}\quad\text{(proved)}$

$\,3(iii)\,$ Prove that $\,\,\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}-2\cos\frac{9\pi}{13}\cos\frac{12\pi}{13}=0$

Sol. $\,\,\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}-2\cos\frac{9\pi}{13}\cos\frac{12\pi}{13}\\=2\cos\frac{4\pi}{13}\cos\frac{\pi}{13}-2\cos\frac{9\pi}{13}\cos\frac{12\pi}{13}\\=2\cos\frac{4\pi}{13}\cos\frac{\pi}{13}-2\cos\left(\pi-\frac{4\pi}{13}\right)\cos\left(\pi-\frac{\pi}{13}\right)\\=2\cos\frac{4\pi}{13}\cos\frac{\pi}{13}-2\left(-\cos\frac{4\pi}{13}\right)\left(-\cos\frac{\pi}{13}\right)\\=0\quad\text{(proved)}$

$\,3(iv)\,$ Prove that $\,\,4\sin15^{\circ}\sin75^{\circ}=\sqrt2(\cos105^{\circ}+\sin75^{\circ})$

Sol. $\,\text{L.H.S.}=4\sin15^{\circ}\sin75^{\circ}\\=2\times [2\sin75^{\circ}\sin15^{\circ}]\\=2\times [\cos(75^{\circ}-15^{\circ})-\cos(75^{\circ}+15^{\circ})]\\=2\times [\cos60^{\circ}-\cos90^{\circ}]\\=2\times \frac 12\\=1$

$\,\,\text{R.H.S.}=\sqrt2(\cos105^{\circ}+\sin75^{\circ})\\=\sqrt2\left[\cos(90^{\circ}+15^{\circ})+\sin75^{\circ}\right]\\=\sqrt2[-\sin15^{\circ}+\sin75^{\circ}]\\=\sqrt2 \times 2\cos\frac{75^{\circ}+15^{\circ}}{2}\sin\frac{75^{\circ}-15^{\circ}}{2}\\=\sqrt2 \times 2\cos45^{\circ}\sin30^{\circ}\\=\sqrt2\times2\times \frac{1}{\sqrt2}\times \frac 12\\=1$

Hence, L.H.S.=R.H.S. (proved)

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$\,4(i)\,$ Prove that $\,\,\sin80^{\circ}\cos20^{\circ}+\sin45^{\circ}\cos145^{\circ}\\+\sin55^{\circ}\cos245^{\circ}=0$

Solution.

$\,\,\sin80^{\circ}\cos20^{\circ}+\sin45^{\circ}\cos145^{\circ}\\+\sin55^{\circ}\cos245^{\circ}\\=\frac 12[2\sin80^{\circ}\cos20^{\circ}+2\sin45^{\circ}\cos145^{\circ}\\+2\sin55^{\circ}\cos245^{\circ}]\\=\frac 12[(\sin(80^{\circ}+20^{\circ})+\sin(80^{\circ}-20^{\circ}))\\+(\sin(145^{\circ}+45^{\circ})-\sin(145^{\circ}-45^{\circ}))\\+(\sin(245^{\circ}+55^{\circ})-\sin(245^{\circ}-55^{\circ}))]\\=\frac 12[\sin100^{\circ}+\sin60^{\circ}+\sin190^{\circ}-\sin100^{\circ}\\+\sin300^{\circ}-\sin190^{\circ}]\\=\frac 12[\sin60^{\circ}+\sin(360^{\circ}-60^{\circ})]\\=\frac 12[\sin60^{\circ}-\sin60^{\circ}]\\=0\quad\text{(proved)}$

$\,4(ii)\,$ Prove that, $\,\cos32^{\circ}\sin20^{\circ}+\cos144^{\circ}\cos2^{\circ}\\+\sin68^{\circ}\cos56^{\circ}=0$

Solution.

$\,\,\cos32^{\circ}\sin20^{\circ}+\cos144^{\circ}\cos2^{\circ}\\+\sin68^{\circ}\cos56^{\circ}\\=\frac 12\left(2\cos32^{\circ}\sin20+2\cos144^{\circ}\cos2^{\circ}\\+2\sin68^{\circ}\cos56^{\circ}\right)\\=\frac 12\left[\sin(32^{\circ}+20^{\circ})-\sin(32^{\circ}-20^{\circ})\\+\cos(144^{\circ}+2^{\circ})+\cos(144^{\circ}-2^{\circ})\\+\sin(68^{\circ}+56^{\circ})+\sin(68^{\circ}-56^{\circ})\right]\\=\frac 12\left[\sin52^{\circ}-\sin12^{\circ}\\+\cos146^{\circ}+\cos142^{\circ}+\sin124^{\circ}+\sin12^{\circ}\right]\\=\frac 12\left[\sin52^{\circ}+\cos(90^{\circ}+56^{\circ})\\+\cos(90^{\circ}+52^{\circ})+\sin(180^{\circ}-56^{\circ})\right]\\=\frac 12[\sin52^{\circ}-\sin56^{\circ}-\sin52^{\circ}+\sin56^{\circ}]\\=0\,\,\,\text{(proved)}$

$\,4(iii)\,$ Prove that, $\,\,\cos306^{\circ}+\cos234^{\circ}+\cos162^{\circ}+\cos18^{\circ}=0$

Solution.

$\,\,\cos306^{\circ}+\cos234^{\circ}+\cos162^{\circ}+\cos18^{\circ}\\=2\cos\frac{306^{\circ}+234^{\circ}}{2}\cos\frac{306^{\circ}-234^{\circ}}{2}\\+2\cos\frac{162^{\circ}+18^{\circ}}{2}\cos\frac{162^{\circ}-18^{\circ}}{2}\\=2\cos270^{\circ}\cos36^{\circ}+\cos90^{\circ}\cos72^{\circ}\\=2\cos(3\times 90^{\circ}+0^{\circ})\cos36^{\circ}+0\,\,[*]\\=-2\sin0^{\circ}\cos36^{\circ}\\=0\,\,\text{(proved)}$

Note [*]:  $\,\,\cos 90^{\circ}=0=\sin0^{\circ}$

$\,4(iv)\,$ Prove that, $\,\,\cos10^{\circ}\cos20^{\circ}+\sin45^{\circ}\cos145^{\circ}\\+\sin55^{\circ}\cos245^{\circ}=0$

Solution.

$\,\,\cos10^{\circ}\cos20^{\circ}+\sin45^{\circ}\cos145^{\circ}\\+\sin55^{\circ}\cos245^{\circ}\\=\frac12\left(2\cos10^{\circ}\cos20^{\circ}+2\sin45^{\circ}\cos145^{\circ}\\+2\sin55^{\circ}\cos245^{\circ}\right)\\=\frac 12\left[\cos(20^{\circ}+10^{\circ})+\cos(20^{\circ}-10^{\circ})\\+\sin(145^{\circ}+45^{\circ})-\sin(145^{\circ}-45^{\circ})\\+\sin(245^{\circ}+55^{\circ})-\sin(245^{\circ}-55^{\circ})\right]\\=\frac 12\left[\cos30^{\circ}+\cos10^{\circ}+\sin190^{\circ}-\sin100^{\circ}\\+\sin300^{\circ}-\sin190^{\circ}\right]\\=\frac 12[\cos30^{\circ}+\cos10^{\circ}-\sin(90^{\circ}+10^{\circ})\\+\sin(360^{\circ}-60^{\circ})]\\=\frac 12[\cos30^{\circ}+\cos10^{\circ}-\cos10^{\circ}-\sin60^{\circ}]\\=\frac 12\left(\frac{\sqrt3}{2}-\frac{\sqrt3}{2}\right)\\=0\,\,\text{(proved)}$

$\,4(v)\,$ Prove that, $\,\,\cos24^{\circ}+\cos55^{\circ}+\cos125^{\circ}+\cos204^{\circ}\\+\cos300^{\circ}=\frac 12.$

Sol. $\,\,\cos24^{\circ}+\cos55^{\circ}+\cos125^{\circ}+\cos204^{\circ}\\+\cos300^{\circ}\\=(\cos55^{\circ}+\cos125^{\circ})+(\cos24^{\circ}+\cos204^{\circ})\\+\cos300^{\circ}\\=2\cos\frac{125^{\circ}+55^{\circ}}{2}\cos\frac{125^{\circ}-55^{\circ}}{2}+2\cos\frac{204^{\circ}+24^{\circ}}{2} \\ \times \cos\frac{204^{\circ}-24^{\circ}}{2}+\cos(360^{\circ}-60^{\circ})\\=2\cos90^{\circ}\cos35^{\circ}+2\cos114^{\circ}\cos90^{\circ} \\+\cos60^{\circ}\\=0+0+\frac 12\\=\frac 12\,\,\,\text{(proved)}$

$\,5(i)\,$ Prove that, $\,\,\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}=\frac 18$

Sol. $\,\,\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}\\=\frac 12 \cos20^{\circ}[2\cos40^{\circ}\cos80^{\circ}]\\=\frac 12\cos20^{\circ}[\cos(80^{\circ}+40^{\circ})+\cos(80^{\circ}-40^{\circ})]\\=\frac 12\cos20^{\circ}[\cos120^{\circ}+\cos40^{\circ}]\\=\frac12\cos20^{\circ}\cos(90^{\circ}+30^{\circ}) \\ +\frac 14 \times 2\cos20^{\circ}\cos40^{\circ}\\=-\frac 12\cos20^{\circ}\sin30^{\circ}+\frac 14[\cos(40^{\circ}+20^{\circ})\\+\cos(40^{\circ}-20^{\circ})]\\=-\frac 14\cos20^{\circ}+\frac 14[\cos60^{\circ}+\cos20^{\circ}]\\=-\frac 14\cos20^{\circ}+\frac 14\times \frac 12+\frac 14\cos20^{\circ}\\=\frac 18\,\,\text{(proved)}$

$\,5(ii)\,$ Prove that, $\,\,8\sin20^{\circ}\sin40^{\circ}\sin80^{\circ}=\sqrt3$

Sol. $\,\,8\sin20^{\circ}\sin40^{\circ}\sin80^{\circ}\\=4\sin20^{\circ}[2\sin40^{\circ}\sin80^{\circ}]\\=4\sin20^{\circ}[\cos(80^{\circ}-40^{\circ})-\cos(80^{\circ}+40^{\circ})]\\=4\sin20^{\circ}[\cos40^{\circ}-\cos120^{\circ}]\\=4\sin20^{\circ}[\cos40^{\circ}-\cos(90^{\circ}+30^{\circ})]\\=4\sin20^{\circ}[\cos40^{\circ}+\sin30^{\circ}]\\=4\sin20^{\circ}\cos40^{\circ}+4\sin20^{\circ}\sin30^{\circ}\\=2(2\cos40^{\circ}\sin20^{\circ})+4\sin20^{\circ}\times \frac 12\\=2[\sin(40^{\circ}+20^{\circ})-\sin(40^{\circ}-20^{\circ})]\\+2\sin20^{\circ}\\=2[\sin60^{\circ}-\sin20^{\circ}]+2\sin20^{\circ}\\=2.\frac{\sqrt3}{2}-2\sin20^{\circ}+2\sin20^{\circ}\\=\sqrt3\,\,\,\text{(proved)}$

$\,5(iii)\,$ Prove that, $\,\,\tan20^{\circ}\tan40^{\circ}\tan80^{\circ}=\sqrt3$

Sol. $\,\,\tan20^{\circ}\tan40^{\circ}\tan80^{\circ}\\=\frac{\sin20^{\circ}\sin40^{\circ}\sin80^{\circ}}{\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}}\\=\frac{\frac{\sqrt3}{8}}{\frac 18}\,\,[*]\\=\sqrt3\,\,\text{(proved)}$

Note[*] : Use the value of $\,\sin20^{\circ}\sin40^{\circ}\sin80^{\circ}\,$ and $\,\cos20^{\circ}\cos40^{\circ}\cos80^{\circ}\,$ from $\,\,5(i,ii).$

$\,5(iv)\,$ Prove that, $\,\cos40^{\circ}\cos100^{\circ}\cos160^{\circ}=\frac18$

Sol. $\,\cos40^{\circ}\cos100^{\circ}\cos160^{\circ}\\=\frac 12\cos40^{\circ}\left[2\cos100^{\circ}\cos160^{\circ}\right]\\=\frac 12\cos40^{\circ}[\cos(160^{\circ}+100^{\circ})\\+\cos(160^{\circ}-100^{\circ})]\\=\frac12\cos40^{\circ}[\cos260^{\circ}+\cos60^{\circ}]\\=\frac 14 \times 2\cos40^{\circ}\cos260^{\circ}+\frac12\cos40^{\circ}\cos60^{\circ}\\=\frac14[\cos(260^{\circ}+40^{\circ})+\cos(260^{\circ}-40^{\circ})]\\+\frac12\cos40^{\circ}\times \frac12\\=\frac14[\cos300^{\circ}+\cos220^{\circ}]+\frac14\cos40^{\circ}\\=\frac14[\cos(360^{\circ}-60^{\circ})\\+\cos(180^{\circ}+40^{\circ})]+\frac14\cos40^{\circ}\\=\frac14\cos60^{\circ}-\frac14\cos40^{\circ}+\frac14\cos40^{\circ}\\=\frac14\times \frac12\\=\frac18$

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