Ako vypočítate hriech (cos ^ -1 (5/13) + tan ^ -1 (3/4))?

odpoveď:

#sin (cos ^ (- 1) (5/13) + tan ^ (- 1) (3/4)) = 63/65 #

vysvetlenie:

nechať #cos ^ (- 1) (5/13) = x # potom

# Rarrcosx = 5/13 #

# Rarrsinx = sqrt (1-cos ^ 2 x) = sqrt (1- (5/13) ^ 2) = 12/13 #

# Rarrx = sin ^ (- 1) (12/13) = cos ^ (- 1) (5/13) #

Tiež, nech #tan ^ (- 1) (3/4) = y # potom

# Rarrtany = 3/4 #

# Rarrsiny = 1 / cscy = 1 / sqrt (1 + postieľka ^ 2y) = 1 / sqrt (1+ (4/3) ^ 2) = 3/5 #

# Rarr = tan ^ (- 1) (3/4) = sin ^ (- 1) (3/5) #

#rarrcos ^ (- 1) (5/13) + tan ^ (- 1) (3/4) #

# = Sin ^ (- 1) (12/13) + sin ^ (- 1) (3/5) #

# = Sin ^ (- 1) (12/13 * sqrt (1- (3/5) ^ 2) + 3/5 * sqrt (1- (12/13) ^ 2)) #

# = Sin ^ (- 1) (12/13 * 4/5 + 3/5 * 5/13) = 63/65 #

teraz, #sin (cos ^ (- 1) (5/13) + tan ^ (- 1) (3/4)) #

# = Sin (sin ^ (- 1) (63/65)) = 63/65 #

Ako vypočítate hriech (2sin ^ -1 (10x))?

Sin (2sin ^ (- 1) (10x)) = 20xsqrt (1-100x ^ 2) "Nech" y = hriech (2sin ^ (- 1) (10x)) Teraz, "" theta = sin ^ (- 1 ) (10x) "" => sin (theta) = 10x => y = sin (2theta) = 2sinthetacostheta Pripomeňme, že: "" cos ^ 2theta = 1-sin ^ 2theta => costheta = sqrt (1-sin ^ 2theta) => y = 2sinthetasqrt (1-sin ^ 2theta) => y = 2 * (10x) sqrt (1- (10x) ^ 2) = farba (modrá) (20xsqrt (1-100x ^ 2))

Ako vypočítate hriech ((13pi) / 6)?

Sin ((13pi) / 6) = sin (2pi + pi / 6) = sin (pi / 6) = 1/2

Preukázať, že Cot 4x (hriech 5 x + hriech 3 x) = Cot x (hriech 5 x - hriech 3 x)?

# sin a + sin b = 2 sin ((a + b) / 2) cos ((ab) / 2) sin a - sin b = 2 sin ((ab) / 2) cos ((a + b) / 2 ) Pravá strana: postieľka x (sin 5x - sin 3x) = postieľka 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 Ľavá strana: postieľka (4x) (sin 5x + sin 3x) = postieľka (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 Sú rovnaké quad sqrt #