Suppose X is a rotation angle such that 90degrees<x<180degrees and cosx=-3/5. find the trig function values for an angle of rotation that is half as big (x/2)?

Question

sin(x/2)=
cos(x/2)=
tan(x/2)=

please help I am so confused and don't even know where to begin.

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Answer 1 · 2018-08-12T05:09:01

# 2/sqrt5, 1/sqrt5 and 2.#

In # ( 90^o, 180^o ) cos x = cos ( 126.87^o) = - 3/ 5#,

So, x/2 = 63.44*o in Q_1 ( 0, 90^o ), wherein all are positive.

And so,

#sin (x/2) = sqrt (1/2 ( 1 - cos x )) = sqrt (1/2 ( 1 + 3/5 ))= 2/sqrt5#

#cos (x/2) = sqrt (1/2 ( 1 + cos x )) = sqrt (1/2 ( 1 - 3/5 ))#

#=1/sqrt(5)#

#tan (x/2) = # the ratio #(2/sqrt3)/(1/sqrt5) = 2#