Question

4(cos³10+sin³20)=3(cos10+sin20) prove this question

Answer 1

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 ) Equation.
4(cos3 10° + sin3 20°) = 3(cos10° + sin20°) + cos30° - sin60°
4(cos3 10° + sin3 20°) = 3(cos10° + sin20°) [as we know that cos30=sin60]

Answer 2

Answer:

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Step-by-step explanation:

Let us consider the LHS

4(cos3 10° + sin3 20°)

As we know that, sin 60° = √3/2 = cos 30°

Sin 30° = cos 60° = 1/2

Therefore,

Sin (3×20°) = cos (3×10°)

3sin 20° – 4sin320° = 4cos310° – 3cos 10°

(As we know, sin 3θ = 3sin θ – 4sin3 θ and cos 3θ = 4cos3θ – 3cosθ)

Therefore,

4(cos310° + sin320°) = 3(sin 20° + cos 10°)

= RHS

Thus proved.