Question

If theta = 30°, show that cos6^theta + sin^6theta = 1-3 sin2^theta cos2^theta​

Answer 1

Answer:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\theta = 30 \degree \:$

Consider

$\rm :\longmapsto\: {cos}^{6}\theta + {sin}^{6}\theta$

On substituting the value, we get

$\rm \:  =  \: {cos}^{6}30\degree + {sin}^{6}30\degree$

$\rm \:  =  \: {(cos30\degree)}^{6} + {(sin30\degree)}^{6}$

$\rm \:  =  \: {\bigg[\dfrac{ \sqrt{3} }{2} \bigg]}^{6} + {\bigg[\dfrac{1}{2} \bigg]}^{6}$

$\rm \:  =  \: \dfrac{27}{64} + \dfrac{1}{64}$

$\rm \:  =  \: \dfrac{27 + 1}{64}$

$\rm \:  =  \: \dfrac{28}{64}$

$\rm \:  =  \: \dfrac{7}{16}$

Now, Consider

$\rm :\longmapsto\:1 - 3 {sin}^{2}\theta {cos}^{2}\theta$

On substituting the value, we get

$\rm \:  =  \: 1 - 3 {sin}^{2}30\degree {cos}^{2}30\degree$

$\rm \:  =  \: 1 - 3 \times {\bigg[\dfrac{ \sqrt{3} }{2} \bigg]}^{2} \times {\bigg[\dfrac{1}{2} \bigg]}^{2}$

$\rm \:  =  \: 1 - 3 \times \dfrac{3}{4} \times \dfrac{1}{4}$

$\rm \:  =  \: 1 - \dfrac{9}{16}$

$\rm \:  =  \: \dfrac{16 - 9}{16}$

$\rm \:  =  \: \dfrac{7}{16}$

Hence, We concluded that

$\rm\implies \:\boxed{\tt{ {sin}^{6}\theta + {cos}^{6}\theta = 1 - 3 {sin}^{2}\theta {cos}^{2}\theta }}$

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$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$