Given cos x = 1/4, with x in the fourth quadrant, what is the exact value of sin x/2?

1 Answer
Shubh
Mar 29, 2018

#sin (x/2)#=#sqrt6/4#

Explanation:

as, cos x =#1/4#=#(base)/"hypotenuse"#
so perpendicular=#sqrt5#
also, #(3pi)/2#< x<2#pi#
so, divide by 2 will give,
#(3pi)/4#<#x/2#<#pi#
therefore sin#x/2# is (+) in II quadrant
now, cos x =1-2#sin^2(x/2)#
#sin^2(x/2)#=#(1-cosx)/2#
#sin^2(x/2)#=#(1-1/4)/2#
#sin^2(x/2)#=#3/8#
#sin(x/2)#=#(sqrt3)/sqrt8#=#sqrt3/(2sqrt2)#
on rationalising,
#sin(x/2)=sqrt6/4#

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