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

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

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Answer 1 · 2018-03-29T16:01:34

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

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

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Explanation:

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