I am confused to use which trigonometric formula
when a long question comes
4(cos^3 10^° + sin^3 20^°)=3(cos 10^° + sin 20^°)

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Solution

cos 3θ = 4cos³ θ - 3cosθ
4cos³ θ = cos3θ + 3 cosθ
Taking θ = 10°
4cos³ 10° = cos30° + 3cos10° ... ( 1 )

sin3θ = 3sinθ - 4sin³ θ
4sin³ θ = 3sinθ - sin3θ
Taking θ = 20°
4sin³ 20° = 3sin20° - sin60° ... ( 2 )

Adding ( 1 ) and ( 2 ),
4(cos³ 10° + sin³ 20°) = 3(cos10° + sin20°) + cos30° - sin60°
4(cos³ 10° + sin³ 20°) = 3(cos10° + sin20°) [as we know that cos30=sin60]

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