Cos - 1 (sqrtcos α) - tan -1 (sqrtcos α) = x이면, sin x의 값은 얼마인가?

대답:

# sinx = tan (α / 2) -cosalpha / (sqrt2cos (α / 2)) #

설명:

방해 # sqrtcosalpha = m #

# rarrcos ^ (- 1) (m) -tan ^ (- 1) (m) = x #

방해 #cos ^ (- 1) m = y # 그때 # cozy = m #

# rarrsiny = sqrt (1-cos ^ 2y) = sqrt (1-m ^ 2) #

(1-m ^ 2)) = cos ^ (- 1) m #

또한, #tan ^ (- 1) m = z # 그때 # tanz = m #

# rarrsinz = 1 / cscz = 1 / sqrt (1 + cot ^ 2z) = 1 / sqrt (1+ (1 / m) ^ 2) = m / sqrt (1 + m ^ 2) #

(1 + m ^ 2)) = tan ^ (- 1) m #

# rarrcos ^ (- 1) (m) -tan ^ (- 1) (m) #

(1 + m ^ 2)) - sin ^ (- 1) (m / sqrt (1 + m ^ 2)

sqrt (1 - m ^ 2)) * sqrt (1 - m ^ 2) - (m / sqrt (1 + m ^ 2) (1- (sqrt (1-m ^ 2)) ^ 2)) #

cosalpha / sqrt (1 + cosalpha)) # = sin ^ (- 1) (sqrt (1-cosalpha) / (1 + cosalpha)

sinα / (sqrt2cos (α / 2))) = x # sin (-1) (tan (α /

tan (α / 2) -cosalpha / (sqrt2cos (α / 2)) # rarrsinx = sin (sin ^ (-1) (tan (α / 2) - cosα / (sqrt2cos