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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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