• Matéria: Matemática
  • Autor: jhenniferfga
  • Perguntado 9 anos atrás

Transforme em produto as expressões:
a) sen (2x + y) + sen (2x - y)
b) cos (a+b+c) + cos (a+b-c)

Respostas

respondido por: Niiya
4
sen~p+sen~q=2*sen~\left(\dfrac{p+q}{2}\right)*cos~\left(\dfrac{p-q}{2}\right)\\\\\\cos~p+cos~q=2*cos~\left(\dfrac{p+q}{2}\right)*cos~\left(\dfrac{p-q}{2}\right)
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a)

sen~(2x+y)+sen~(2x-y)=2\cdot sen~(\frac{2x+y+2x-y}{2})\cdot cos~(\frac{2x+y-(2x-y)}{2})\\\\sen~(2x+y)+sen~(2x-y)=2\cdot sen~(\frac{4x}{2})\cdot cos~(\frac{2y}{2})\\\\\boxed{\boxed{sen~(2x+y)+sen~(2x-y)=2\cdot sen~(2x)\cdot cos~y}}

b)

cos~(a+b+c)+cos~(a+b-c)=cos~([a+b]+c)+cos~([a+b]-c)\\\\cos~(a+b+c)+cos~(a+b-c)=2\cdot cos~(\frac{a+b+c+a+b-c}{2})\cdot cos~(\frac{a+b+c-(a+b-c)}{2})\\\\cos~(a+b+c)+cos~(a+b-c)=2\cdot cos~(\frac{2a+2b}{2})\cdot cos~(\frac{2c}{c})\\\\\boxed{\boxed{cos~(a+b+c)+cos~(a+b-c)=2\cdot cos~(a+b)\cdot cos~c}}
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