Given #sin(θ) = (sqrt(5 - sqrt(5)))/5# and #cos(θ) > 0# how do you use identities to find the exact functional value of #tan theta#?

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Answer 1 · 2015-06-17T18:19:06

From #sin^2 theta + cos^2 theta = 1# and #cos theta > 0# we get:

#cos theta = sqrt(1 - sin^2 theta) = sqrt(20+sqrt(5))/5#

#tan theta = sin theta / cos theta = #

#sin^2 alpha + cos^2 alpha = 1# for all #alpha in RR#

We are given #cos theta > 0#

So:

#cos theta = sqrt(1 - sin^2 theta)#

#=sqrt(1-(sqrt(5-sqrt(5))/5)^2)#

#=sqrt(1-(5-sqrt(5))/25)#

#=sqrt((25-(5-sqrt(5)))/25)#

#=sqrt(20+sqrt(5))/5#

#tan theta = sin theta / cos theta#

#=(sqrt(5-sqrt(5))/5)/(sqrt(20+sqrt(5))/5)#

#=sqrt(5-sqrt(5))/sqrt(20+sqrt(5))#

#=sqrt((5-sqrt(5))/(20+sqrt(5)))#

#=sqrt((5-sqrt(5))/(20+sqrt(5))*(20-sqrt(5))/(20-sqrt(5)))#

#=sqrt((105-25sqrt(5))/395)#

#=sqrt((21-5sqrt(5))/79)#