Question #1e28e

Question

1161 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-15T08:47:49

see below

$tan3theta=tan(2theta+theta)$

#using the compound angle identity

$tan(x+y)=(tanx+tany)/(1-tanxtany)$

$tan3theta=(tan2theta+tantheta)/(1-tan2thetatantheta)$

using the same identity on $" "tan2theta$

$tan3theta=((tantheta+tantheta)/(1-tanthetatantheta)+tantheta)/(1-(tantheta+tantheta)/(1-tanthetatantheta)tantheta)$

now to tidy up

$tan3theta=((2tantheta)/(1-tan^2theta)+tantheta)/(1-(2tantheta)/(1-tan^2theta)tantheta)$

$tan3theta=((2tantheta+tantheta-tan^3theta)/cancel((1-tan^2theta)))/((1-tan^2theta-2tan^2theta)/cancel((1-tan^2theta)))$

$tan3theta=((2tantheta+tantheta-tan^3theta))/((1-tan^2theta- 2tan^2theta)$

giving

$tan3theta=(3tantheta-tan^3theta)/(1-3tan^2theta)$