Question 622815
There is a formula that expresses tan(2u) in terms of tan(u):
{{{tan(2u) = (2*tan(u))/(1-tan^2(u))}}}
So if we can find tan(u) we can use this formula to find tan(2u).<br>
We are given that sin(u) = -4/5. We can use one of the Pythagorean identities, {{{cos^2(a) = 1- sin^2(a)}}}, to use sin(u) to find cos(u). And once we have cos(u) we can use the fact that tan(u) = sin(u)/cos(u) to find tan(u).<br>
{{{cos^2(u) = 1 - sin^2(u)}}}
{{{cos^2(u) = 1 - (-4/5)^2}}}
{{{cos^2(u) = 1 - 16/25}}}
{{{cos^2(u) = 25/25 - 16/25}}}
{{{cos^2(u) = 9/25}}}
Now we find the square root of each side. But which square root should we use? The positive one or the negative one? Well, since we are told that {{{pi< u < 3pi/2}}} we know that u terminates in the 3rd quadrant. And in the 3rd quadrant cos is negative. So we should use the negative square root:
{{{cos(u) = -sqrt(9/25)}}}
{{{cos(u) = -3/5}}}<br>
Now we can find tan(u):
{{{tan(u) = sin(u)/cos(u)}}}
{{{tan(u) = (-4/5)/(-3/5)}}}
{{{tan(u) = 4/3}}}<br>
Now we can find tan(2u):
{{{tan(2u) = (2*tan(u))/(1-tan^2(u))}}}
{{{tan(2u) = (2*(4/3))/(1-(4/3)^2)}}}
{{{tan(2u) = (8/3)/(1-(16/9))}}}
{{{tan(2u) = (8/3)/(9/9-(16/9))}}}
{{{tan(2u) = (8/3)/(-7/9)}}}
When we divide fractions we change it to multiplying by the reciprocal:
{{{tan(2u) = (8/3)*(-9/7)}}}
{{{tan(2u) = -24/63}}}
which reduces to:
{{{tan(2u) = -8/21}}}