Hvordan beregner du synd (cos ^ -1 (5/13) + tan ^ -1 (3/4))?

Svar:

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

Forklaring:

La #cos ^ (- 1) (5/13) = x # deretter

# 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) #

Også, la #tan ^ (- 1) (3/4) = y # deretter

# Rarrtany = 3/4 #

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

# Rarry = 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 #

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

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

Hvordan beregner du synd (2sin ^ -1 (10x))?

Synd (2sin ^ (- 1) (10x)) = 20xsqrt (1-100x ^ 2) "La" y = sin (2sin ^ (- 1) (10x)) La nå "" teta = sin ^ ) (10x) "" => synd (theta) = 10x => y = sin (2theta) = 2sinthetacostheta Husk at: "" cos ^ 2theta = 1-sin ^ 2theta => costheta = sqrt (1-sin ^ 2theta) => y = 2sinthetasqrt (1-sin ^ 2theta) => y = 2 * (10x) sqrt (1- (10x) ^ 2) = farge (blå) (20xsqrt (1-100x ^ 2))

Hvordan beregner du synd ((13pi) / 6)?

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

Bevis: - synd (7 theta) + synd (5 theta) / synd (7 theta) -sin (5 theta) =?

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