How do you simplify #tan4theta# to trigonometric functions of a unit #theta#?

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Answer 1 · 2016-09-26T10:54:42

#tan4theta=(4tantheta-4tan^3theta)/(1-6tan^2theta+tan^4theta)#

Using the identity #tan2theta=(2tantheta)/(1-tan^2theta)#

#tan4theta=(2tan2theta)/(1-tan^2(2theta)#

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

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

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

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

= #(4tantheta-4tan^3theta)/(1-6tan^2theta+tan^4theta)#