Using #(1-cos2x)/(1+cos2x)=tan^2x#, how do you find the exact value of #tan(pi/8)#?
2 Answers
Shown below...
Explanation:
Let
Rationalise:
You can also simplify this further if you know how to:
Both answers are the same!
Explanation:
#tan^2(pi/8)=(1-cos(pi/4))/(1+cos(pi/4)#
#color(white)(xxxxxxx)=(1-1/sqrt2)/(1+1/sqrt2)#
#color(white)(xxxxxxx)=(sqrt2-1)/(sqrt2+1)xx(sqrt2-1)/(sqrt2-1)#
#color(white)(xxxxxxx)=(sqrt2-1)^2#
#"since "pi/8" is acute then angle in first quadrant"#
#tan(pi/8)=sqrt2-1#