How do you find the exact functional value tan(-5pi/12) using the cosine sum or difference identity?

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Answer 1 · 2015-09-07T09:17:04

#tan((-5pi)/12)=-2-sqrt(3)#

First #tan# in an odd function, so you can calculate #tan((5pi)/12)# and then multiply calculated value by #-1#.

To calculate #tan((5pi)/12)# you can write it as:

#tan((5pi)/12)=tan((2pi)/12+(3pi)/12)=tan(pi/6+pi/4)#

Now you can use the sum identity:

#tan(alpha+beta)=(tan alpha+tan beta)/(1-tanalpha*tanbeta)#

So you get:

#tan((5pi)/12)=(tan(pi/6)+tan(pi/4))/(1-tan(pi/6)*tan(pi/4))=#

#=(sqrt(3)/3+1)/(1-sqrt(3)/3)=(1+sqrt(3)/3)^2/((1-sqrt(3)/3)*(1+sqrt(3)/3))=#

#=(1+(2sqrt(3))/3+1/3)/(1-1/3)=(4/3+(2sqrt(3))/3)/(2/3)=#

#=3/2*(4/3+(2sqrt(3))/3)=3/2*(4+2sqrt(3))/3=2+sqrt(3)#

Finally if #tan((5pi)/12)=2+sqrt(3)#, then #tan((-5pi)/12)=-2-sqrt(3)#