Question

Sendo m= sen + cos, usando produtos notáveis, calcular:
a)sen³ + cos³
b)seb⁴ + cos⁴

por favor rápido é urgente alguém me ajuda por favor


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Answer 1

Resposta:

Explicação passo-a-passo:

Como m=senx+cosx

x=α

A) sen³ x+ cos³ x

elevamos ao cubo em ambos lados

$(senx+cos x)^3=m^3\\sen^3x + 3 * sen^2x * cos x + 3 * sen x * cos^2x + cos^3x=m^3\\sen^3x + 3 * sen x * cos x (sen x + cos x) + cos^3x = m^3\\sen^3x + 3 * sen x * cos x * m + cos^3x = m^3\\sen^3x + cos^3x=m^3- 3m\frac{(m^2- 1)}{2}\\sen^3x + cos^3x = m^3 -\frac{(3m^3- 3m)}{2}sen^3x + cos^3x = \frac{(2m^3- 3m^3 + 3m)}{2}\\sen^3x + cos^3x = \frac{(- m^3+ 3m)}{2}$

B) sen⁴x + cos⁴x

sen²x+cos²x=1  elevamos ao quadrado ambos lados

$(sen^2x+cos^2x)^2=1^2\\(sen^2x)^2+2*sen^2x*cos^2x+ (cos^2x)^2=1\\sen^4x+cos^4x=1-2*sen^2x*cos^2x$

Answer 2

Resposta:

Explicação passo-a-passo:

a)

$m=sen\alpha+cos\alpha\\\\m^{2}=(sen\alpha+cos\alpha)^{2}\\ \\m^{2}=sen^{2}\alpha+cos^{2}\alpha+2.sen\alpha.cos\alpha\\\\m^{2}=1+2.sen\alpha.cos\alpha\\\\2.sen\alpha.cos\alpha=m^{2}-1\\\\sen\alpha.cos\alpha=\dfrac{m^{2}-1}{2}$

$m=sen\alpha+cos\alpha\\\\m^{3}=(sen\alpha+cos\alpha)^{3}\\\\m^{3}=sen^{3}\alpha+cos^{3}\alpha+3.sen\alpha.cos\alpha.(sen\alpha+cos\alpha)\\\\sen^{3}\alpha+cos^{3}\alpha=m^{3}-3.sen\alpha.cos\alpha.(sen\alpha+cos\alpha)\\\\sen^{3}\alpha+cos^{3}\alpha=m^{3}-3.\dfrac{(m^{2}-1)}{2}.m\\\\sen^{3}\alpha+cos^{3}\alpha=\dfrac{2m^{3}-3m.(m^{2}-1)}{2}\\\\sen^{3}\alpha+cos^{3}\alpha=\dfrac{2m^{3}-3m^{3}+3m}{2}\\\\sen^{3}\alpha+cos^{3}\alpha=\dfrac{3m-m^{3}}{2}$

b)

$sen\alpha+cos\alpha=m\\\\(sen\alpha+cos\alpha)^{4}=m^{4}\\\\sen^{4}\alpha+4.sen^{3}.cos\alpha+6.sen^{2}\alpha.cos^{2}\alpha+4.sen\alpha.cos^{3}+cos^{4}\alpha=m^{4}\\\\sen^{4}\alpha +cos^{4}\alpha=m^{4}-4.sen^{3}.cos\alpha-4.sen\alpha.cos^{3}-6.sen^{2}\alpha.cos^{2}\alpha\\\\sen^{4}\alpha +cos^{4}\alpha=m^{4}-4.sen\alpha.cos\alpha(sen^{2}\alpha +cos^{2}\alpha)-6.(sen\alpha.cos\alpha)^{2}\\\\sen^{4}\alpha +cos^{4}\alpha=m^{4}-4.\dfrac{m^{2}-1}{2}.1-6.(\dfrac{m^{2}-1}{2})^{2}$

$sen^{4}\alpha +cos^{4}\alpha=m^{4}-2.(m^{2}-1})-6.(\dfrac{m^{4}-2m^{2}+ 1}{4})\\\\sen^{4}\alpha +cos^{4}\alpha=m^{4}-2m^{2}-2-3.(\dfrac{m^{4}-2m^{2}+ 1}{2})\\\\sen^{4}\alpha +cos^{4}\alpha=\dfrac{2m^{4}-4m^{2}-4-3m^{4}+6m^{2}-3}{2}\\\\sen^{4}\alpha +cos^{4}\alpha=\dfrac{-m^{4}+2m^{2}-7}{2}$