Kako preverite? Tan x + cos x = sin x (sec x + cotan x)

Odgovor:

Glej spodaj.

Pojasnilo:

# LHS = tanx + cosx #

# = sinx / cosx + cosx #

# = sinx (1 / cosx + cosx / sinx) #

# = sinx (secx + cotx) = RHS #

Odgovor:

Glej dokazilo spodaj

Pojasnilo:

Potrebujemo

# secx = 1 / cosx #

# tanx = sinx / cosx #

# cotx = cosx / sinx #

Zato, # LHS = tanx + cosx #

# = sinx / cosx + cosx #

# = sinx (1 / cosx + cosx / sinx) #

# = sinx (secx + cotx) #

# = RHS #

# QED #

Kako preverite 2 (tan (2A)) * (2 (cos ^ 2 (2A) - sin ^ 2 (4A)) = sin (8A)?

Prikazano pod 2tan (2A) xx2 [cos ^ 2 (2A) -sin ^ 2 (4A)] = sin (8A) LHS = leva in RHS = desna stran. Zato začnem z leve strani in pokažem, da je enaka desni strani. LHS = 2tan (2A) xx [2cos ^ 2 (2A) -2sin ^ 2 (4A)] = 4tan (2A) cos ^ 2 (2A) -4tan2Asin ^ 2 (4A) = 4 (sin (2A)) / cos (2A) cos ^ 2 (2A) -4 (sin (2A)) / cos (2A) sin ^ 2 (4A) = 4sin (2A) cos (2A) -4 (sin (2A)) / cos (2A) sin ^ 2 (2 (2A)) = 2 * 2sin (2A) cos (2A) -4 (sin (2A)) / cos (2A) xx2sin ^ 2 (2A) cos ^ 2 (2A) = 2sin (2 ( 2A)) - 4 (sin (2A)) xx2sin ^ 2 (2A) cos (2A) = 2sin (4A) -4 * 2sin (2A) cos (2A) xxsin ^ 2 (2A) = 2sin (4A) -4sin (4A) sin ^ 2 (2A) = 2sin

Kako preverite [sin ^ 3 (B) + cos ^ 3 (B)] / [sin (B) + cos (B)] = 1-sin (B) cos (B)?

Dokaz pod Ekspanzijo ^ 3 + b ^ 3 = (a + b) (^ 2-ab + b ^ 2) in lahko uporabimo to: (sin ^ 3B + cos ^ 3B) / (sinB + cosB) = ((sinB + cosB) (sin ^ 2B-sinBcosB + cos ^ 2B)) / (sinB + cosB) = sin ^ 2B-sinBcosB + cos ^ 2B = sin ^ 2B + cos ^ 2B-sinBcosB (identiteta: sin ^ 2x + cos ^ 2x = 1) = 1-sinBcosB

Kako preverite posteljico (x) / sin (x) -tan (x) / cos (x) = csc (x) sec (x) 1 / (sin (x) + cos (x))?

"To ni res, zato izpolnite x = 10 ° npr. In videli boste, da enakost ne drži." "Nič več dodati."