Using (1-cos2x)/(1+cos2x)=tan^2x, how do you find the exact value of tan(pi/8)?

2 Answers
May 29, 2018

Shown below...

Explanation:

Let x = pi/ 8

=> tan^2 (pi/8) = (1 - cos(2 * pi/ 8) ) / ( 1 + cos(2 * pi/ 8) )

=> tan^2(pi/8) = ( 1 - sqrt(2)/2 ) / ( 1 + sqrt(2)/2 )

= ( 2/2 - sqrt(2)/2 ) / ( 2/2 + sqrt(2)/2 )

= (( 2-sqrt(2) ) / 2) / (( 2 + sqrt(2) ) /2 )

= ( 2 - sqrt(2) ) / ( 2 + sqrt(2) )

Rationalise:

= ( 2 - sqrt(2) ) / ( 2 + sqrt(2) ) * ( 2 - sqrt(2) ) / ( 2 - sqrt(2) )

= ( 6 -4sqrt(2 ) ) / ( 2 )

=> tan^2 (pi/8) = 3 - 2sqrt(2)

color(red)(=> tan ( pi/8) = sqrt( 3 - 2 sqrt(2) )

You can also simplify this further if you know how to:

color(red)( => tan(pi/8) = sqrt(2) -1

Both answers are the same!

May 29, 2018

tan(pi/8)=sqrt2-1

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