Question

senx +cos x =m calcule sen ao cubo de x + cos ao cubo x

Answer 1

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Definamos

$\large\begin{array}{l}\mathsf{sen\,x=a,\;\;cos\,x=b}\\\\ \textsf{de modo que }\mathsf{a+b=m.} \end{array}$


$\large\begin{array}{l} \textsf{Retomemos a expans\~ao dos bin\^omios:}\\\\ \mathsf{a^3+3a^2b+3ab^2+b^3=(a+b)^3}\\\\ \mathsf{a^3+b^3=(a+b)^3-(3a^2b+3ab^2)}\\\\ \mathsf{a^3+b^3=(a+b)^3-3ab(a+b)\qquad(i)} \end{array}$


$\large\begin{array}{l} \textsf{Mas temos tamb\'em que}\\\\ \mathsf{a^2+2ab+b^2=(a+b)^2}\\\\ \mathsf{2ab=(a+b)^2-(a^2+b^2)\qquad\quad\textsf{(mas }}\mathsf{a^2+b^2=1}\textsf{)}\\\\ \mathsf{2ab=(a+b)^2-1}\\\\ \mathsf{ab=\dfrac{\,1\,}{2}\big[(a+b)^2-1\big]}\\\\ \mathsf{3ab=\dfrac{\,3\,}{2}\big[(a+b)^2-1\big]\qquad(ii)} \end{array}$


$\large\begin{array}{l} \textsf{Substituindo (ii) em (i), temos}\\\\ \mathsf{a^3+b^3=(a+b)^3-\dfrac{\,3\,}{2}\big[(a+b)^2-1\big]\cdot (a+b)}\\\\ \mathsf{a^3+b^3=m^3-\dfrac{\,3\,}{2}\,(m^2-1)\cdot m}\\\\ \mathsf{a^3+b^3=m^3-\dfrac{\,3\,}{2}\,(m^3-m)}\\\\ \mathsf{a^3+b^3=m^3-\dfrac{\,3\,}{2}\,m^3+\dfrac{\,3\,}{2}\,m} \end{array}$

$\large\begin{array}{l} \mathsf{a^3+b^3=-\,\dfrac{\,1\,}{2}\,m^3+\dfrac{\,3\,}{2}\,m}\\\\ \mathsf{a^3+b^3=-\,\dfrac{\,1\,}{2}\,m\cdot (m^2-3)}\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathsf{sen^3\,x+cos^3\,x=-\,\dfrac{\,1\,}{2}\,m\cdot (m^2-3)} \end{array}}\qquad\quad\checkmark \end{array}$


$\large\textsf{Bons estudos! :-)}$


Tags:   identidade transformação trigonométrica fatoração soma de potência seno cosseno tangente cotangente sen cos tg tan cotg cot binômio de newton triângulo de pascal trigonometria