Question 821433
{{{90 < beta < 180 }}}
This tells us that {{{beta}}} terminates in the second quadrant, where x-coordinates are negative and y-coordinates are positive.<br>
At this point a diagram may be helpful:<ul><li>Draw an angle in standard position which terminates in the 2nd quadrant. This angle will be {{{beta}}}.</li><li>From somewhere on the terminal side draw a perpendicular down to the x-axis. This perpendicular, the x-axis and the terminal side form a right triangle.</li><li>We've been given that {{{tan(beta) = -sqrt(5)/2}}}. Since tan is opposite/adjacent, we want to label the opposite site with the numerator of {{{-sqrt(5)/2}}} and the adjacent side with the denominator. But where does the "=" go? Answer: Since we are in the 2nd quadrant, the minus should go with the adjacent side (x-axis). So label the opposite side, the perpendicular, as {{{sqrt(5)}}} and label the adjacent side (x-axis) as -2.</li></ul>For {{{tan(beta/2)}}} we will use the tan((1/2)x) identity:
{{{tan((1/2)x) = sin(x)/(1+cos(x))}}}
From this we can see that we will need {{{sin(beta)}}} and {{{cos(beta)}}}. For both sin and cos we need the hypotenuse. So we use the Pythagorean Theorem to find the hypotenuse:
{{{(-2)^2+(sqrt(5))^2 = h^2}}} (where "h" represents the hypotenuse)
Simplifying...
{{{4+5 = h^2}}}
{{{9 = h^2}}}
{{{3 = h}}}  (ignoring the negative square root of 9 since hypotenuse's are never negative).<br>
Now that we have the hypotenuse, we can find the sin, opposite/hypotenuse, and the cos, adjacent/hypotenuse, of {{{beta}}}:
{{{sin(beta) = sqrt(5)/3}}}
{{{cos(beta) = (-2)/3}}}<br>
And finally we can find {{{tan(beta/2)}}}:
{{{tan(beta/2) = sin(beta)/(1+cos(beta))}}}
Substituting in the values we found for sin and cos:
{{{tan(beta/2) = (sqrt(5)/3)/(1+((-2)/3))}}}
Simplifying...
{{{tan(beta/2) = (sqrt(5)/3)/(3/3+((-2)/3))}}}
{{{tan(beta/2) = (sqrt(5)/3)/(1/3)}}}
{{{tan(beta/2) = ((sqrt(5)/3)/(1/3))(3/3)}}}
{{{tan(beta/2) = sqrt(5)/1}}}
{{{tan(beta/2) = sqrt(5)}}}