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

1 Answer
Jun 5, 2018

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