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Si senx + cosx = 1 / 2 hallar k = cos3x - sen3x?

Si senx + cosx = 1 / 2 hallar k = cos3x - sen3x.

Mejor respuesta

Fabianmh2001
10

Senx + cosx = 1 / 2

k = cos3x - sen3x

k = cons3x - sen3x

k = cos(2x + x) - sen(2x + x)

k = cos2x cosx - sen2x senx - sen2x cosx - cos2x senx

k = cos2x cosx - cos2x senx - sen2x senx - sen2x cosx

k = cos2x (cosx - senx) - sen2x (senx + cosx)

k = (cos ^ 2 x - sen ^ 2 x)(cosx - senx) - 2senx cosx (senx + cosx)

k = (cosx + senx)(cosx - senx) (cosx - senx) - 2 senx cosx (senx + cosx)

k = (cosx + senx) [(cosx - senx)(cosx - senx) - 2 cosxsenx]

k = (cosx + senx) (cos ^ 2 x - 2cosx + sen ^ 2 - 2 cosx senx)

k = (cosx + senx) (1 - 4 cosx senx)

k = (1 / 2) (1 - 4 cosx senx)

k = (1 / 2) - 2 cosx senx

senx + cosx = 1 / 2 Elevas al cuadrado ambos lados para no alterar

(senx + cosx) ^ 2 = (1 / 2) ^ 2

sen ^ 2 x + 2senx cosx + cos ^ 2 x = 1 / 4

1 + 2 senx cosx = 1 / 4

2 senx cosx = (1 / 4) - 1

2 senx cosx = - 3 / 4

k = (1 / 2) - ( - 3 / 4)

k = (1 / 2) + (3 / 4)

k = 5 / 4.

Otras 1 respuestas

Respuesta 2

Pizza7
6

<img src="https://tex.z-dn.net/?f=K%3D%5Ccos%203x-%5Csin%203x%5C%5C%20%5C%5C%0AK%3D%284%5Ccos%5E3x-3%5Ccos%20x%29-%283%5Csin%20x-4%5Csin%5E3x%29%5C%5C%20%5C%5C%0AK%3D4%28%5Ccos%5E3x%2B%5Csin%5E3x%29-3%28%5Csin%20x%2B%5Ccos%20x%29%5C%5C%20%5C%5C%0AK%3D4%28%5Ccos%20x%2B%5Csin%20x%29%28%5Ccos%5E2x-%5Csin%20x%5Ccos%20x%2B%5Csin%5E2%20x%29-%5Cdfrac%7B3%7D%7B2%7D%5C%5C%20%5C%5C%0AK%3D2%281-%5Csin%20x%5Ccos%20x%29-%5Cdfrac%7B3%7D%7B2%7D%5C%5C%20%5C%5C" />

Por otra parte

<img src="https://tex.z-dn.net/?f=%5Csin%20x%2B%5Ccos%20x%20%3D%201%2F2%5C%5C%20%5C%5C%0A%28%5Csin%20x%2B%5Ccos%20x%29%5E2%3D1%2F4%5C%5C%20%5C%5C%0A1%2B2%5Csin%20x%5Ccos%20x%3D1%2F4%5C%5C%20%5C%5C%0A-%5Csin%20x%5Ccos%20x%20%3D%203%2F8" />

entonces

<img src="https://tex.z-dn.net/?f=K%3D2%281%2B3%2F8%29-3%2F2%5C%5C%20%5C%5C%0A%5Cboxed%7BK%3D%5Cdfrac%7B5%7D%7B4%7D%7D" />.