Cos ¹ (sqrtcos α) tan ¹ (sqrtcos α) = x, akkor mi a sin x értéke?

Válasz:

# Sinx = tan (alfa / 2) -cosalpha / (sqrt2cos (alfa / 2)) #

Magyarázat:

enged # Sqrtcosalpha = m #

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

enged #cos ^ (- 1) m = y # azután # Otthonos = m #

# Rarrsiny = sqrt (1-cos ^ 2Y) = sqrt (1-m ^ 2) #

# Rarry = sin ^ (- 1) (sqrt (1-m ^ 2)) = cos ^ (- 1) m #

Továbbá hagyja #tan ^ (- 1) m = z # azután # Tanz = m #

# Rarrsinz = 1 / cscz = 1 / sqrt (1 + gyermekágy ^ 2Z) = 1 / sqrt (1+ (1 / m) ^ 2) = m / sqrt (1 + m ^ 2) #

# Rarrz = sin ^ (- 1) (m / sqrt (1 + m ^ 2)) = tan ^ (- 1) m #

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

# = Sin ^ (- 1) (sqrt (1-m ^ 2)) - sin ^ (- 1) (m / sqrt (1 + m ^ 2)) #

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

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

# = Sin ^ (- 1) (tan (alfa / 2) -cosalpha / (sqrt2cos (alfa / 2))) = X #

# Rarrsinx = sin (sin ^ (- 1) (tan (alfa / 2) -cosalpha / (sqrt2cos (alfa / 2)))) = tan (alfa / 2) -cosalpha / (sqrt2cos (alfa / 2)) #