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Question Number 101116 by Jamshidbek2311 last updated on 30/Jun/20

Answered by 1549442205 last updated on 07/Jul/20

$$\mathbf{We}\mathbf{prove}\mathbf{that}\mathbf{tan}\frac{3\mathit{\pi }}{11}+4\mathbf{sin}\frac{2\mathit{\pi }}{11}=\sqrt{11}$$$$\iff \sin \frac{3\pi }{11}+4\sin \frac{2\pi }{11}\cos \frac{3\pi }{11}=\sqrt{11}\cos \frac{3\pi }{11}$$$$\iff 3\sin \frac{\pi }{11}-{4\sin }^3\frac{\pi }{11}+4(2\sin \frac{\pi }{11}\cos \frac{\pi }{11}-\sqrt{11})({4\cos }^3\frac{\pi }{11}-3\cos \frac{\pi }{11})=0$$$$\iff \sin \frac{\pi }{11}[3-4(1-{\cos }^2\frac{\pi }{11})]+4(2\sin \frac{\pi }{11}\cos \frac{\pi }{11}-\sqrt{11})({4\cos }^3\frac{\pi }{11}-3\cos \frac{\pi }{11})=0$$$$\iff \sin \frac{\pi }{11}({4\cos }^2\frac{\pi }{11}-1)+4(2\sin \frac{\pi }{11}\cos \frac{\pi }{11}-\sqrt{11})({4\cos }^3\frac{\pi }{11}-3\cos \frac{\pi }{11})=0$$$$\mathrm{Putting}\cos \frac{\pi }{11}=\mathrm{x}\Rightarrow \sin \frac{\pi }{11}=\sqrt{1-{\mathrm{x}}^2}\mathrm{we}\mathrm{get}$$$$\sqrt{1-{\mathbf{x}}^2}(4{\mathbf{x}}^2-1)+4(2\mathbf{x}\sqrt{1-{\mathbf{x}}^2}-\sqrt{11})(4{\mathbf{x}}^3-3\mathbf{x})=0$$$$\iff \sqrt{1-{\mathrm{x}}^2}({32\mathrm{x}}^4-{20\mathrm{x}}^2-1)=\sqrt{11}({4\mathrm{x}}^3-3\mathrm{x})$$$$\mathrm{Square}\mathrm{two}\mathrm{sides}\mathrm{of}\mathrm{above}\mathrm{we}\mathrm{get}$$$$(1-{\mathrm{x}}^2)({1024\mathrm{x}}^8-{1280\mathrm{x}}^6+{336\mathrm{x}}^4+{40\mathrm{x}}^2+1)=11({16\mathrm{x}}^6-{24\mathrm{x}}^4+{9\mathrm{x}}^2)$$$$\iff (-{1024\mathrm{x}}^{10}+{2304\mathrm{x}}^8-{1616\mathrm{x}}^6+{296\mathrm{x}}^4+{39\mathrm{x}}^2+1={176\mathrm{x}}^6-{264\mathrm{x}}^4+{99\mathrm{x}}^2$$$$\iff {1024\mathrm{x}}^{10}-{2304\mathrm{x}}^8+{1792\mathrm{x}}^6-{560\mathrm{x}}^4+{60\mathrm{x}}^2-1=0$$$$=({32\mathrm{x}}^5-{16\mathrm{x}}^4-{32\mathrm{x}}^3+{12\mathrm{x}}^2+6\mathrm{x}-1)({32\mathrm{x}}^5+{16\mathrm{x}}^4-{32\mathrm{x}}^3-{12\mathrm{x}}^2+6\mathrm{x}+1)=0$$$$\iff (32{\mathbf{x}}^5-16{\mathbf{x}}^4-32{\mathbf{x}}^3+12{\mathbf{x}}^2+6\mathbf{x}-1)=0(\ast )$$$$(\mathrm{because}({32\mathrm{x}}^5+{16\mathrm{x}}^4-{32\mathrm{x}}^3-{12\mathrm{x}}^2+6\mathrm{x}+1\ne 0\mathrm{for}\mathrm{x}=\frac{\pi }{11})$$$$\mathbf{Now}\mathbf{we}\mathbf{need}\mathbf{prove}\mathbf{the}\mathbf{equality}\mathbf{is}\mathbf{true}.$$$$\mathbf{Indeed},$$$$\mathrm{Applying}{\mathrm{Mauvra}}^{\prime }\mathrm{s}\mathrm{formular}\mathrm{we}\mathrm{have}$$$${(\cos \frac{\pi }{11}+\mathrm{isin}\frac{\pi }{11})}^5=\cos \frac{5\pi }{11}+\mathrm{isin}\frac{5\pi }{11}$$$$\iff {\cos }^5\frac{\pi }{11}+{5\mathrm{icos}}^4\frac{\pi }{11}\sin \frac{\pi }{11}+{10\mathrm{i}}^2{\cos }^3\frac{\pi }{11}{\sin }^2\frac{\pi }{11}+{10\cos }^2.{\mathrm{i}}^3{\sin }^3\frac{\pi }{11}+5\cos \frac{\pi }{11}.{\mathrm{i}}^4{\sin }^4\frac{\pi }{11}+{\sin }^5\frac{\pi }{11}=\cos \frac{5\pi }{11}+\mathrm{isin}\frac{5\pi }{11}$$$$\Rightarrow \mathbf{cos}\frac{5\mathit{\pi }}{11}={\cos }^5\frac{\pi }{11}-{10\cos }^3\frac{\pi }{11}{\sin }^2\frac{\pi }{11}+5\cos \frac{\pi }{11}{\sin }^4\frac{\pi }{11}$$$$={\cos }^5\frac{\pi }{11}-{10\cos }^3\frac{\pi }{11}(1-{\cos }^2\frac{\pi }{11})+5\cos \frac{\pi }{11}{(1-{\cos }^2\frac{\pi }{11})}^2$$$$=16{\mathbf{cos}}^5\frac{\mathit{\pi }}{11}-20{\mathbf{cos}}^3\frac{\mathit{\pi }}{11}+5\mathbf{cos}\frac{\mathit{\pi }}{11}\left(1\right)$$$${(\cos \frac{\pi }{11}+\mathrm{isin}\frac{\pi }{11})}^6=\cos \frac{6\pi }{11}+\mathrm{isin}\frac{6\pi }{11}=-\cos \frac{5\pi }{11}+\mathrm{isin}\frac{5\pi }{11}$$$${\cos }^6\frac{\pi }{11}+{6\cos }^5\frac{\pi }{11}.\mathrm{isin}\frac{\pi }{11}+{15\cos }^4\frac{\pi }{11}.{\mathrm{i}}^2{\sin }^2\frac{\pi }{11}+{20\cos }^3\frac{\pi }{11}.{\mathrm{i}}^3{\sin }^3\frac{\pi }{11}+{15\cos }^2\frac{\pi }{11}.{\mathrm{i}}^4{\sin }^4\frac{\pi }{11}$$$$+5\cos \frac{\pi }{11}.{\mathrm{i}}^5{\sin }^5\frac{\pi }{11}+{\sin }^6\frac{\pi }{11}=-\cos \frac{5\pi }{11}+\mathrm{isin}\frac{5\pi }{11}$$$$\iff {\mathbf{x}}^6-15{\mathbf{x}}^4{\mathbf{y}}^2+15{\mathbf{x}}^2{\mathbf{y}}^4+{\mathbf{y}}^6+\mathbf{i}.(6{\mathbf{x}}^5\mathbf{y}-15{\mathbf{x}}^4{\mathbf{y}}^2-20{\mathbf{x}}^3{\mathbf{y}}^3)=-\mathbf{cos}\frac{5\mathit{\pi }}{11}+\mathbf{isin}\frac{5\mathit{\pi }}{11}(\mathbf{y}=\mathbf{sin}\frac{\mathit{\pi }}{11})\left(2\right)$$$$\mathrm{From}\left(1\right)\left(2\right)\mathrm{we}\mathrm{get}$$$${\mathrm{x}}^6-{15\mathrm{x}}^4{\mathrm{y}}^2+{15\mathrm{x}}^2{\mathrm{y}}^4+{\mathrm{y}}^6=-{16\mathrm{x}}^5+{20\mathrm{x}}^3-5\mathrm{x}$$$$\iff {\mathrm{x}}^6-{15\mathrm{x}}^4(1-{\mathrm{x}}^2)+{15\mathrm{x}}^2{(1-{\mathrm{x}}^2)}^2+{(1-{\mathrm{x}}^2)}^3=-{16\mathrm{x}}^5+{20\mathrm{x}}^3-5\mathrm{x}$$$$\iff {32\mathrm{x}}^6-{48\mathrm{x}}^4+{18\mathrm{x}}^2-1=-{16\mathrm{x}}^5+{20\mathrm{x}}^3-5\mathrm{x}$$$$\iff {32\mathrm{x}}^6+{16\mathrm{x}}^6-{48\mathrm{x}}^4-{20\mathrm{x}}^3+{18\mathrm{x}}^2+5\mathrm{x}-1=0$$$$\iff (\mathrm{x}+1)({16\mathrm{x}}^5-{16\mathrm{x}}^4-{32\mathrm{x}}^3+{12\mathrm{x}}^2+6\mathrm{x}-1)=0$$$$\iff 16{\mathbf{x}}^5-16{\mathbf{x}}^4-32{\mathbf{x}}^3+12{\mathbf{x}}^2+6\mathbf{x}-1=0$$$$\mathbf{Thus},\mathbf{the}\mathbf{equality}(\ast )\mathbf{proved}$$$$\mathbf{Therefore},\mathbf{tan}\frac{3\mathit{\pi }}{11}+4\mathbf{sin}\frac{2\mathit{\pi }}{11}=\sqrt{11}(\mathbf{q}.\mathbf{e}.\mathbf{d})$$

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