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Question
Prove that: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]
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Solution
\[\text{ We know, } \]
\[ \sin60 °= \cos30 ° \left( = \frac{\sqrt{3}}{2} \right)\]
\[ \Rightarrow \sin3 \times 20 ° = \cos3 \times 10 °\]
\[ \Rightarrow 3\sin20 °- 4 \sin^3 20 °= 4 \cos^3 10 °- 3\cos10 ° \]
\[ \left( \because \sin3\theta = 3sin\theta - 4 \sin^3 \theta \text{ and } \cos3\theta = 4 \cos^3 \theta - 3cos\theta \right) \]
\[ \Rightarrow 4\left( \cos^3 10 + \sin^3 20 ° \right) = 3\left( \cos10°+ \sin20 ° \right)\]
\[\text{ Hence proved } .\]
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