How do you find the exact value of #tan2theta=4tantheta# in the interval #0<=theta<2pi#?

Question

2246 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-02T08:29:33

#theta={0,0.6155,(pi-0.6155),pi,(pi+0.6155),(2pi-0.6155)}#

#tan2theta=4tantheta# can be written as

#(2tantheta)/(1-tan^2theta)=4tantheta#

or #2tantheta=4tantheta(1-tan^2theta)#

or #2tantheta-4tantheta(1-tan^2theta)=0#

or #2tantheta(1-2(1-tan^2theta))=0#

or #2tantheta(1-2+2tan^2theta)=0#

or #2tantheta(2tan^2theta-1)=0#

or #2tantheta(sqrt2tantheta-1)(sqrt2tantheta+1)=0#

i.e. #tantheta=0=tan0# or #1/sqrt2=tan0.6155# or #-1/sqrt2=tan(-0.6155)#, where #+-0.6155# is angle in radians and is equal to #+-35.5^@#

and hence in range #0 <= theta < 2pi#

#theta={0,0.6155,(pi-0.6155),pi,(pi+0.6155),(2pi-0.6155)}#
graph{tan(2x)-4tanx [-1, 9, -2.5, 2.5]}