How do you find #tan(1/2sin^-1(1/2))#?

Question

3255 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-05T14:32:27

#rarrtan(1/2(sin^(-1)(1/2))#

#=tan(30^@/2) or tan(150^@/2)#

#=tan15^@ or tan75^@#

Now, #rarrtan15^@#

#=tan(45^@-30^@)#

#=(tan45^@-tan30^@)/(1+tan30^@tan45^@)#

#=(1-(1/sqrt3))/(1+1/sqrt3)#

#=(sqrt3-1)/(sqrt3+1)xx(sqrt3-1)/(sqrt3-1)#

#=((sqrt(3))^2-2sqrt3+1^2)/((sqrt3)^2-1^2)=(4-2sqrt3)/2=2-sqrt3#

Similary, #rarrtan75^@#

#=tan(45^@+30^@)#

#=(tan45^@+tan30^@)/(1-tan30^@tan45^@)#

#=(1+(1/sqrt3))/(1-1/sqrt3)#

#=(sqrt3+1)/(sqrt3-1)xx(sqrt3+1)/(sqrt3+1)#

#=((sqrt(3))^2+2sqrt3+1^2)/((sqrt3)^2-1^2)=(4+2sqrt3)/2=2+sqrt3#