What is the value of #tan^2(1/2 sin^(-1)(2/3))# ?

2 Answers
George C.
Jan 4, 2018

#tan^2(1/2 sin^(-1)(2/3)) = (7-3sqrt(5))/2#

Explanation:

Note that:

#tan (alpha+beta) = (tan alpha + tan beta)/(1-tan alpha tan beta)#

So:

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

So:

#(tan 2 theta) tan^2 theta + 2tan theta - (tan 2theta) = 0#

So:

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

#color(white)(tan theta) = (-1+-sqrt(1+tan^2 2 theta))/(tan 2 theta)#

#color(white)(tan theta) = (-1+-sec 2 theta)/(tan 2 theta)#

Note that if:

#sin(2 theta) = 2/3#

Then:

#cos(2 theta) = +-sqrt(1-sin^2 2 theta) = +-sqrt(1-4/9) = +-sqrt(5)/3#

In fact if:

#2 theta = sin^(-1) (2/3)#

then #2 theta in [-pi/2, pi/2]#, so #cos 2 theta >= 0# and #2 theta# is in Q1.

So:

#cos 2 theta = sqrt(5)/3#

#sec 2 theta = 1/(cos 2 theta) = 3/sqrt(5) = (3sqrt(5))/5#

#tan 2 theta = (sin 2 theta)/(cos 2 theta) = (2/3) / (sqrt(5)/3) = 2/sqrt(5) = (2sqrt(5))/5#

Then:

#tan theta = (-1+sec 2 theta)/(tan 2 theta) = (-1+(3sqrt(5))/5) / ((2sqrt(5))/5) = (3-sqrt(5))/2#

So:

#tan^2 theta = (3-sqrt(5))^2/4 = (9-6sqrt(5)+5)/4 = (7-3sqrt(5))/2#

turksvids
Jan 4, 2018

#(7-3sqrt(5))/2#

Explanation:

#sin^-1(2/3)# is a QI angle, which means all values of its trig functions are positive.

Let #x = sin^-1(2/3)#.

#tan^2(x/2) = sin^2(x/2)/cos^2(x/2)#.

#sin^2(x/2) = (1-cos(x))/2# and #cos^2(x/2) = (1+cos(x))/2#

So now we have:

#tan^2(x/2) = ((1-cos(x))/2)/((1+cos(x))/2) = (1-cos(x))/(1+cos(x))#

So since #x = sin^-1(2/3)#, we need to find #cos(x) = cos(sin^-1(2/3))#.

If you draw a right triangle in QI with hypotenuse 3 and opposite side 2 you can find the adjacent side is #sqrt(5)#, so:

#cos(x) = cos(sin^-1(2/3)) = sqrt(5)/3#.

Since we found:

#tan^2(x/2) = (1-cos(x))/(1+cos(x))# we can substitute:

#tan^2(x/2) = (1-sqrt(5)/3)/(1+sqrt(5)/3)#

we can simplify this:

#(1-sqrt(5)/3)/(1+sqrt(5)/3) = ((3-sqrt(5))/3)/((3+sqrt(5))/3) = (3-sqrt(5))/(3+sqrt(5))#

You can then rationalize this if you'd like:

#(3-sqrt(5))/(3+sqrt(5))*(3-sqrt(5))/(3-sqrt(5))=(14-6sqrt(5))/4=(7-3sqrt(5))/2#

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