Question 727492
{{{5pi/4=pi+pi/4}}} and since tangent has a period of {{{pi}}}
{{{tan(5pi/4)=tan(pi/4)}}}
{{{pi/4}}} (in radians) is {{{45^o}}} , half of a right angle.
 
You should know (or quickly figure out)
sine, cosine, and tangent of {{{30^o}}} , {{{45^o}}} and {{{60^o}}} ,
because a right triangle with {{{30^o}}} and {{{60^o}}} angles is half of an equilateral triangle,
and a right triangle with two {{{45^o}}} angles (an isosceles right triangle) is half of a square.
 
{{{drawing(300,300,-1,11,-1,11,
green(rectangle(0,0,10,10)),
triangle(0,0,10,0,10,10),locate(5,5,sqrt(2)a),
locate(0.6,1,45^o),locate(4.5,0.7,a),locate(9.5,5.5,a),
rectangle(0,0,0.5,0.5),rectangle(0,10,0.5,9.5),
rectangle(10,0,9.5,0.5),rectangle(10,10,9.5,9.5)
)}}} {{{tan(45^o)=a/a=1}}} and {{{sin(45^o)=cos(45^o)=a/(sqrt(2)a)=1/sqrt(2)=sqrt(2)/2}}}
So {{{tan(5pi/4)=tan(pi/4)=tan(45^o)=1}}}
and you should know that {{{sin(pi/2)=sin(90^o)=1}}}
The angle between 0 and {{{pi}}} {{{(180^o)}}} with a sine of 1 is {{{highlight(pi/2)}}} and that is sin^-1(tan(5pi/4))