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

2 Answers
Rhys
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!

Jim G.
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#

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