Question

# If $$\sin { \theta } +\sin ^{ 2 }{ \theta } =1,$$ then prove that $$\cos ^{ 12 }{ \theta } +3\cos ^{ 10 }{ \theta } +3\cos ^{ 8 }{ \theta } +\cos ^{ 6 }{ \theta } -1=0$$

Solution

## The question should be if $$\sin{\theta}+{\sin}^{2}{\theta}=1$$ then prove that$${\cos}^{12}{\theta}+{\cos}^{6}{\theta}+3{\cos}^{10}{\theta}+3{\cos}^{8}{\theta}-1=0$$We have $$\sin{\theta}+{\sin}^{2}{\theta}=1$$$$\Rightarrow\,\sin{\theta}=1-{\sin}^{2}{\theta}$$$$\Rightarrow\,\sin{\theta}={\cos}^{2}{\theta}$$    ..........    $$[sin²θ+cos²θ=1]$$Squaring both sides, we get$$\Rightarrow\,\sin^2{\theta}={\cos}^{4}{\theta}$$$$\Rightarrow\,1-{\cos}^{2}{\theta}={\cos}^{4}{\theta}$$    ......    $$[sin²θ+cos²θ=1]$$$$\Rightarrow\,{\cos}^{4}{\theta}+{\cos}^{2}{\theta}=1$$Cubing both sides,we get$$\Rightarrow\,{\left({\cos}^{4}{\theta}+{\cos}^{2}{\theta}\right)}^{3}=1$$Using, $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$$$\Rightarrow\,{\cos}^{12}{\theta}+{\cos}^{6}{\theta}+3{\cos}^{6}{\theta}\left({\cos}^{4}{\theta}+{\cos}^{2}{\theta}\right)=1$$$$\Rightarrow\,{\cos}^{12}{\theta}+{\cos}^{6}{\theta}+3{\cos}^{10}{\theta}+3{\cos}^{8}{\theta}-1=0$$$$\Rightarrow\,{\cos}^{12}{\theta}+3{\cos}^{10}{\theta}+3{\cos}^{8}{\theta}+{\cos}^{6}{\theta}-1=0$$Hence proved.Mathematics

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