Question 860242
Given Tan x= 2/5 with &#960; < x < 3&#960;/2 
find sin x/2 cos x/2 tan x/2
sin 2x cos 2x and tan 2x
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Reference right  triangle is in quadrant III in which sin<0, cos<0, tan>0
Hypotenuse=&#8730;(2^2+5^2)=&#8730;(4+25)=&#8730;29
sin(x)=-2/&#8730;29
cos(x)=-5/&#8730;29

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{{{sin(x/2)=sqrt((1-cos(x))/2)=sqrt((1+5/sqrt(29))/2)}}}
{{{cos(x/2)=sqrt((1+cos(x))/2)=sqrt((1-5/sqrt(29))/2)}}}
{{{tan(x/2)=sin(x)/(1+cos(x))=(-2/sqrt(29))/(1-5/sqrt(29))=2/(sqrt(29)-5)}}}
{{{sin(2x)=2sin(x)cos(x)=2*(-2/sqrt(29))*(-5/sqrt(29))=20/29}}}
{{{cos(2x)=cos^2(x)-sin^2(x)=25/29-4/29=21/29}}}
{{{tan(2x)=sin(2x)/cos(2x)=20/21}}}

Note: Let me know if my solution(s) are correct, helpful and understandable.