Ano ang ibig sabihin ng cos (arctan (3)) + kasalanan (arctan (4))?

Sagot:

#cos (arctan (3)) + sin (arctan (4)) = 1 / sqrt (10) + 4 / sqrt (17) #

Paliwanag:

Hayaan # tan ^ -1 (3) = x #

pagkatapos # rarrtanx = 3 #

# rarrsecx = sqrt (1 + tan ^ 2x) = sqrt (1 + 3 ^ 2) = sqrt (10) #

# rarrcosx = 1 / sqrt (10) #

# rarrx = cos ^ (- 1) (1 / sqrt (10)) = tan ^ (- 1) (3) #

Gayundin, hayaan #tan ^ (- 1) (4) = y #

pagkatapos # rarrtany = 4 #

# rarrcoty = 1/4 #

# rarrcscy = sqrt (1 + cot ^ 2y) = sqrt (1+ (1/4) ^ 2) = sqrt (17) / 4 #

# rarrsiny = 4 / sqrt (17) #

# rarry = sin ^ (- 1) (4 / sqrt (17)) = tan ^ (- 1) 4 #

Ngayon, #rarrcos (tan ^ (- 1) (3)) + kasalanan (tan ^ (- 1) tan (4)) #

# sinis (cos ^ -1 (1 / sqrt (10))) + sin (sin ^ (- 1) (4 / sqrt (17))) = 1 / sqrt (10) + 4 / sqrt (17) #

Katunayan: - kasalanan (7 theta) + kasalanan (5 theta) / kasalanan (7 theta) -in (5 theta) =?

(sin7x + sin5x) / (sin7x-sin5x) = tan6x * cotx rarr (sin7x + sin5x) / (sin7x-sin5x) = (2sin ((7x + 5x) / 2) ) / (2sin ((7x-5x) / 2) * cos ((7x + 5x) / 2) = (sin6x * cosx) / (sinx * cos6x) = (tan6x) / tanx = tan6x * cottx

Ang kasalanan theta / x = cos theta / y pagkatapos ay ang kasalanan theta - cos theta =?

Kung ang frac {sinas} {x} = frac {cos theta] {y} pagkatapos sin theta - cos theta = pm frac {x - y} {sqrt {x ^ 2 + y ^ 2}} frac { sin angta} {x} = frac {cos theta} {y} frac { sin theta} { cos theta} = frac {x} {y} tan theta = at katabi y kaya cos theta = frac { pm y} {sqrt {x ^ 2 + y ^ 2} sin angta = tan theta cos theta sin theta - cos theta = tan theta cos theta - cos theta = (x / y -1) sin theta - cos theta = pm frac {x - y } {sqrt {x ^ 2 + y ^ 2}}

Patunayan na ang Cot 4x (kasalanan 5 x + kasalanan 3 x) = Cot x (kasalanan 5 x - kasalanan 3 x)?

# sin isang + sin b = 2 sin ((a + b) / 2) cos ((ab) / 2) sin isang - sin b = 2 sin ((ab) / 2) cos ((a + b) / 2 ) Kanang bahagi: 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 Kaliwa: cot (4x) (kasalanan 5x + kasalanan 3x) = cot (4x) cdot 2 kasalanan ((5x + 3x) / 2) cos ((5x-3x) / 2) = {cos 4x} / {sin 4x} cdot 2 sin 4x cos x = 2 cos x cos 4 x They are equal quad sqrt #