How do you find #sin(x/2)# if #cscx=3#?

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Answer 1 · 2015-05-17T16:47:09

Since csc x=3, sin x= #1/3# and therefore cos x=#sqrt(8/9)#= #(2sqrt2)/3#

Since cosx equals #1- 2sin^2 (x/2)#, hence

#2sin^2 (x/2)= 1-cosx#

= 1- #(2sqrt2)/3# =#(3-2sqrt2)/3# = #(sqrt2 -1)^2 /3#

#sin^2 (x/2)= (sqrt2 -1)^2 /6#

#sin(x/2) = (sqrt2-1)/sqrt6#

Answer 2 · 2015-05-17T16:57:45

Use trig identity:# 1 + cot^2 x = 1/sin^2 x = csc^2 x#

#1/sin^2 x = 9 -> sin^2 x = 1/9 -> sin x = +1/3#

# cos x = 1 - sin^2 x = 1 - 1/9 = 8/9 -> cos x = +(sqrt2)/3#

To find #sin (x/2)# use trig identity: #2sin^2 (x/2) = 1 - cos x.#

#2sin^2 (x/2) = 1 - cos x = 3/3 - (sqrt2)/3 = 0.53#

#sin^2 (x/2) = 0.265 -> sin (x/2) = 0.51#