Cos ¹ (sqrtcos α) tan ¹ (sqrtcos α) = x, o zaman sin x'in değeri nedir?

Cevap:

# SiNx = tan (a / 2) -cosalpha / (sqrt2cos (a / 2)) #

Açıklama:

let # Sqrtcosalpha = m #

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

let #cos ^ (- 1) m = y # sonra # Rahat = m #

# Rarrsiny = sqrt (1 Cos ^ 2y) = sqrt (1-m ^ 2) #

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

Ayrıca, izin #tan ^ (- 1) m = z # sonra # Tanz = m #

# Rarrsinz = 1 / cscz = 1 / sqrt (1 + yatağı ^ 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 (a / 2) -cosalpha / (sqrt2cos (a / 2))) X # =

# Rarrsinx = sin (sin ^ (- 1) (tan (a / 2) -cosalpha / (sqrt2cos (a / 2)))) açık kahve renkli (a / 2) -cosalpha / (sqrt2cos (a / 2)) # =