Jawapan:
Penjelasan:
# "menggunakan" formula tambahan warna "(biru)" untuk dosa "#
# • warna (putih) (x) sin (A + -B) = sinAcosB + -cosAsinB #
#rArrsin (A + B) = sinAcosB + cosAsinB #
#rArrsin (A-B) = sinAcosB-cosAsinB #
#rArrsin (A + B) + dosa (A-B) = 2sinAcosB #
Jawapan:
Ia bukan identiti.
Penjelasan:
Ia bukan identiti.
LS:
RS:
Bukti: - dosa (7 theta) + dosa (5 theta) / dosa (7 theta) -in (5 theta) =?
(sin7x + sin5x) / (sin7x-sin5x) = tan6x * cotx rarr (sin7x + sin5x) / (sin7x-sin5x) = (2sin ((7x + 5x) ) / (2sin ((7 x 5x) / 2) * cos ((7x + 5x) / 2) = (sin6x * cosx) / (sinx * cos6x) = (tan6x) / tanx = tan6x * cottx
Sahkan dosa identiti (α + β) dosa (α - β) =?
Sinergi (alpha-beta) = sin ^ 2alpha-sin ^ 2beta rarrsin (alpha + beta) sin (alpha-beta) = 1/2 [2sin (alpha + beta) (Alpha + beta- (alpha-beta)) - cos (alpha + beta + alpha-beta)] = 1/2 [cos2beta-cos2alpha] = 1/2 [1-2sin ^ 2beta - (1-2sin ^ 2alpha)] = sin ^ 2alpha-sin ^ 2beta
Buktikan bahawa Cot 4x (dosa 5 x + dosa 3 x) = Cot x (sin 5 x - sin 3 x)?
2 sin (+ b) / 2) cos ((ab) / 2) sin a - sin b = 2 sin ((ab) / 2) cos ((a + b) ) Sebelah kanan: cot x (sin 5x - sin 3x) = cot x cdot 2 sin ((5x-3x) / 2) cos ((5x + 3x) / 2) = cos x / sin x cdot 2 sin x cos 4x = 2 cos x cos 4x Left side: cot (4x) (sin 5x + sin 3x) = cot (4x) cdot 2 sin ((5x + 3x) / 2) cos ((5x-3x) / 2) = {cos 4x} / {sin 4x} cdot 2 sin 4x cos x = 2 cos x cos 4 x Mereka sama quad sqrt #