Question 1056788
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Find sin(x/2), cos(x/2), and tan(x/2) from the given information 
tan(x)=1; 0 degrees< x <90 degrees 
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The condition "tan(x)=1; 0° < x < 90° " implies

x = 45°.


Then x/2 = 22°30'.


To calculate &nbsp;sin(22°30'), apply the formula of half argument for sines &nbsp;(see the lesson &nbsp;<A HREF= http://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometric-functions-of-half-argument.lesson>Trigonometric functions of half argument</A> in this site):

sin(22°30') = {{{sqrt((1-cos(pi/4))/2)}}} = {{{sqrt((1-sqrt(2)/2)/2)}}} = {{{sqrt((2-sqrt(2))/4)}}} = {{{sqrt(2-sqrt(2))/2}}}.



Similarly, 

cos(22°30') = {{{sqrt((1+cos(pi/4))/2)}}} = {{{sqrt((1+sqrt(2)/2)/2)}}} = {{{sqrt((2+sqrt(2))/4)}}} = {{{sqrt(2+sqrt(2))/2}}}.


Hence,

tan(22°30') = sin(22°30')/cos(22°30') = {{{sqrt((2-sqrt(2))/(2+sqrt(2)))}}} = {{{sqrt( ((2-sqrt(2))*(2-sqrt(2))) / ((2+sqrt(2))*(2-sqrt(2))) )}}} = {{{sqrt( ((2-sqrt(2))^2)/(2^2-(sqrt(2))^2) )}}} = {{{(2-sqrt(2))/sqrt(2)}}} = {{{sqrt(2)-1}}}.
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See the lesson &nbsp;<A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometric-functions-of-half-argument-Examples.lesson>Trigonometric functions of half argument - Examples</A> in this site.