Question

Considere as expressões trigonométricas abaixo:cos(α+β) = cosα cosβ – senα senβ e sen(α+β) = senα cosβ + senβ cosα.Para calcular o cos2α e o sen2α , basta fazer α=β , e, a partir das expressões trigonométricas, obtêm-se:cos2α = cos (α+α) = cos2α – sen2α e sen2α = sen (α+α) = 2senα cosα.De modo semelhante ao cálculo acima, desenvolva o cos3α e o sen3α.

Answer 1

$\\ Cos(3 \alpha )= Cos(2 \alpha + \alpha ) \\ \\ Cos(2 \alpha + \alpha ) = cos(2 \alpha )*Cos( \alpha )-Sen(2 \alpha )*sen( \alpha ) \\ \\ Cos(2 \alpha + \alpha ) =[Cos^2( \alpha )-Sen^2( \alpha )]cos( \alpha )-[2Sen( \alpha )*Cos( \alpha )*Cos \alpha \\ \\ Cos(2 \alpha + \alpha ) =Cos^3( \alpha )-Sen^2( \alpha )Cos( \alpha )-2Sen( \alpha )Cos^2( \alpha ) \\ \\ Cos(2 \alpha + \alpha )= Cos( \alpha )[Cos( \alpha )^2-sen^2( \alpha )-2Sen( \alpha )Cos( \alpha )]$

$\\ Cos(2 \alpha + \alpha )= Cos( \alpha )[Cos( \alpha )^2-sen^2( \alpha )-2Sen( \alpha )Cos( \alpha )] \\ \\ Cos(2 \alpha + \alpha )= Cos( \alpha )[Cos(2 \alpha )-Sen(2 \alpha )]$

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$\\ Sen(3 \alpha )=Sen(2 \alpha + \alpha ) \\ \\ Sen(2 \alpha + \alpha )= sen(2 \alpha )*Cos( \alpha )+sen( \alpha )*Cos(2 \alpha ) \\ \\ Sen(2 \alpha + \alpha )= 2Sen( \alpha )Cos( \alpha )*Cos( \alpha )+Sen( \alpha )*[Cos^2( \alpha )-sen^2 \alpha \\ \\ Sen(2 \alpha + \alpha ) = 2Sen( \alpha )*Cos^2( \alpha )+sen( \alpha )Cos^2( \alpha )-Sen^3( \alpha ) \\ \\ Sen(2 \alpha + \alpha ) = Sen( \alpha )[2Cos^2( \alpha )+Cos^2( \alpha )-Sen^2( \alpha )]$

$\\ Sen(2 \alpha + \alpha ) = Sen( \alpha )[2Cos^2( \alpha )+Cos^2( \alpha )-Sen^2( \alpha )] \\ \\ Sen(2 \alpha + \alpha ) = Sen( \alpha )[3Cos^2( \alpha )-Sen^2( \alpha )]$