Question

SOCORRO!! Se 2x + y = π, então:

a) sen 2x + sen y = 0
b) tg y = tg 2x
c) sen (π - y) = 2 senx 
d) cos 2x = - cosy
e) cos y = cos²x - sen²x

Answer 1

Fórmulas:

$sen~(a\pm b)=sen~a\cdot cos~b\pm sen~b\cdot cos~a\\\\\\cos~(a\pm b)=cos~a\cdot cos~b\mp sen~b\cdot sen~a\\\\\\tg~(a\pm b)=\dfrac{tg~a\pm tg~b}{1\mp tg~a\cdot tg~b}$
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a)

2x + y = π
y = π - 2x

$sen~2x+sen~y=sen~2x+sen(\pi-2x)\\sen~2x+sen~y=sen~2x+sen~\pi\cdot cos~2x-sen~2x\cdot cos~\pi\\sen~2x+sen~y=sen~2x+0-sen~2x\cdot(-1)\\sen~2x+sen~y=sen~2x+sen~2x\\sen~2x+sen~y=2sen~2x$

Falsa

b)

y = π - 2x

$tg~y=tg~(\pi-2x)\\\\\\tg~y=\dfrac{tg~\pi-tg~2x}{1+tg~\pi\cdot tg~2x}\\\\\\tg~y=\dfrac{0-tg~2x}{1+0\cdot tg~2x}\\\\\\tg~y=\dfrac{-tg~2x}{1}\\\\\\tg~y=-tg~2x$

Falsa

c)

y = π - 2x

$sen~(\pi-y)=sen~(\pi-[\pi-2x])\\sen~(\pi-y)=sen~(\pi-\pi+2x)\\sen~(\pi-y)=sen~2x\\sen~(\pi-y)=2\cdot sen~x\cdot cos~x$

Falsa

d)

2x + π = y
2x = y - π

$cos~2x=cos~(\pi-y)\\cos~2x=cos~\pi\cdot cos~y+sen~\pi\cdot sen~y\\cos~2x=-cos~y+0\cdot sen~y\\cos~2x=-cos~y$

Verdadeira, é a resposta da questão

e)

y = π - 2x

$cos~y=cos~(\pi-2x)\\cos~y=cos~\pi\cdot cos~2x+sen~\pi\cdot sen~2x\\cos~y=-cos~2x+0\cdot sen~2x\\cos~y=-cos~2x$

Falsa