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Prove that: \[\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{4}}^{\mathbf{o}}}~\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~{{\mathbf{6}}^{\mathbf{o}}}~=\text{ }\left( \surd \mathbf{5}\text{ }\text{ }\mathbf{1} \right)/\mathbf{8}\]
Prove that: \[\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{4}}^{\mathbf{o}}}~\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~{{\mathbf{6}}^{\mathbf{o}}}~=\text{ }\left( \surd \mathbf{5}\text{ }\text{ }\mathbf{1} \right)/\mathbf{8}\]
CBSE | Class 11 | Exercise 9.3 | Maths | RD Sharma | Trigonometric Functions at Multiples and Submultiples of an Angle
Prove that: \[\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~\mathbf{2}{{\mathbf{4}}^{\mathbf{o}}}~\text{ }\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}~{{\mathbf{6}}^{\mathbf{o}}}~=\text{ }\left( \surd \mathbf{5}\text{ }\text{ }\mathbf{1} \right)/\mathbf{8}\]

Let us consider LHS:

\[si{{n}^{2}}~{{24}^{o}}~\text{ }si{{n}^{2}}~{{6}^{o}}\]

we know, \[sin\text{ }\left( A\text{ }+\text{ }B \right)\text{ }sin\text{ }\left( A\text{ }\text{ }B \right)\text{ }=\text{ }si{{n}^{2}}A\text{ }\text{ }si{{n}^{2}}B\]

Then the above equation becomes,

\[si{{n}^{2}}~{{24}^{o}}~\text{ }si{{n}^{2}}~{{6}^{o}}~=\text{ }sin\text{ }({{24}^{o}}~+\text{ }{{6}^{o}})\text{ }\text{ }sin\text{ }({{24}^{o}}~\text{ }{{6}^{o}})\]

\[=\text{ }sin\text{ }{{30}^{o}}~\text{ }sin\text{ }{{18}^{o}}\]

\[=\text{ }sin\text{ }{{30}^{o}}~\text{ }\left( \surd 5\text{ }\text{ }1 \right)/4\text{ }[since,\text{ }sin\text{ }{{18}^{o}}~=\text{ }\left( \surd 5\text{ }\text{ }1 \right)/4]\]

\[=\text{ }1/2\text{ }\times \text{ }\left( \surd 5\text{ }\text{ }1 \right)/4\]

\[=\text{ }\left( \surd 5\text{ }\text{ }1 \right)/8\]

= RHS

Hence proved.

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