Cos ¹ (sqrtcos α) tan ¹ (sqrtcos α) = x, kakšna je vrednost sin x?

Odgovor:

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

Pojasnilo:

Let # sqrtcosalpha = m #

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

Let #cos ^ (- 1) m = y # potem # cosy = m #

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

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

Tudi, pustite #tan ^ (- 1) m = z # potem # tanz = m #

# rarrsinz = 1 / cscz = 1 / sqrt (1 + cot ^ 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 (alpha / 2))) = x #

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