Question 1204627
If

{{{sin(x) = -20/29}}}and {{{x }}}is in quadrant III, with {{{0 <= x < 360}}}°, find the exact values of the expressions without solving for {{{x}}}.


in Q III both sin and cos are {{{negative}}}

we also know that {{{sin^2(x)+ cos^2(x)=1}}}


solve for {{{cos(x)}}}


{{{cos(x)=sqrt(1-sin^2(x))}}}


substitute given value of {{{sin(x)}}}


{{{cos(x)=sqrt(1-(-20/29)^2)}}}

{{{cos(x)=sqrt(1-400/841)}}}

{{{cos(x)=sqrt(441/841)}}}

{{{cos(x)=21/29}}} or {{{cos(x)=-21/29}}}

we need {{{cos(x)=-21/29}}}



(a) use Half-Angle Identities

{{{sin(x/2)=sqrt((1-cos(x))/2)}}}
​
 {{{sin(x/2)=sqrt((1-(-21/29))/2)}}}

{{{sin(x/2)=sqrt((50/29)/2)}}}

{{{sin(x/2)=sqrt(25/29)}}}

{{{sin(x/2)=(5sqrt(29))/29}}} or {{{sin(x/2)=-(5sqrt(29))/29}}}

in Q III

{{{sin(x/2)=-(5sqrt(29))/29}}}



(b)


{{{cos(x/2)=sqrt((1+cos(x))/2)}}}

{{{cos(x/2)=sqrt((1+(-21/29))/2)}}}

{{{cos(x/2)=sqrt(4/29)}}}

{{{cos(x/2)=(2sqrt(29))/29}}} or {{{cos(x/2)=-(2sqrt(29))/29}}} 

in Q III

{{{cos(x/2)=-(2sqrt(29))/29}}}


(c)

{{{tan(x/2)=(1-cos(x))/(1+cos(x))}}}

{{{tan(x/2)=sqrt((1-(-21/29))/(1+(-21/29)))}}}

{{{tan(x/2)=sqrt(25/4)}}}

{{{tan(x/2)=5/2}}} or {{{tan(x/2)=-5/2}}} 

In the third quadrant, the values for tan are {{{positive}}} only.

{{{tan(x/2)= 5/2}}}