Bagaimana anda mengira dosa (2sin ^ -1 (10x))?

Jawapan:

#sin (2sin ^ (- 1) (10x)) = 20xsqrt (1-100x ^ 2) #

Penjelasan:

# "Let" y = sin (2sin ^ (- 1) (10x)) #

Sekarang, mari # "" theta = sin ^ (- 1) (10x) "" => sin (theta) = 10x #

# => y = sin (2theta) = 2sinthetacostheta #

Ingat bahawa: # "" cos ^ 2theta = 1-sin ^ 2theta => costheta = sqrt (1-sin ^ 2theta) #

# => y = 2sinthetasqrt (1-sin ^ 2theta) #

# => y = 2 * (10x) sqrt (1- (10x) ^ 2) = warna (biru) (20xsqrt (1-100x ^ 2)

Bagaimana anda mengira dosa ((13pi) / 6)?

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

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

Bagaimana anda mengira dosa (cos ^ -1 (5/13) + tan ^ -1 (3/4))?

(5/13) + tan ^ (- 1) (3/4)) = 63/65 Biarkan cos ^ (- 1) (5/13) = x maka rarrcosx = 5/13 (1 / cos ^ 2x) = sqrt (1- (5/13) ^ 2) = 12/13 rarrx = sin ^ (- 1) (12/13) 13) Juga, mari tan ^ (- 1) (3/4) = y maka rarrtany = 3/4 rarrsiny = 1 / cscy = 1 / sqrt (1 + cot ^ 2y) 3) ^ 2) = 3/5 rarry = tan ^ (- 1) (3/4) = sin ^ (- 1) (3/5) 1) (3/4) = sin ^ (- 1) (12/13) + sin ^ (- 1) (3/5) = sin ^ 5) ^ 2) + 3/5 * sqrt (1- (12/13) ^ 2)) = sin ^ (- 1) (12/13 * 4/5 + 3/5 * 65 Sekarang, sin (cos ^ (- 1) (5/13) + tan ^ (- 1) (3/4)) = dosa (sin ^ (- 1) (63/65)) = 63/65