Question

Find the value of sin2a, if tana=-12/5 in the fourth quadrant

Answer 1

$\tan a=-\dfrac{12}5$

Recall the following identities:

$1+\tan^2a=\sec^2a=\dfrac1{\cos^2a}$
$\cos^2a=\dfrac{1+\cos2a}a$

from which we get

$1+\left(-\dfrac{12}5\right)^2=\dfrac1{\cos^2a}$
$\implies\cos^2a=\dfrac{25}{169}$

Since $a$ is in the fourth quadrant $\left(\dfrac{3\pi}2$ we know that $\cos a$ should be positive, so when we take the square root here, we should take the positive root.

$\implies\cos a=\sqrt{\dfrac{25}{169}}=\dfrac5{13}$

Now recall that

$\cos^2a+\sin^2a=1$

and since $a$ is in the fourth quadrant, we expect $\sin a$ to be negative. So,

$\sin^2a=1-\cos^2a=\dfrac{144}{169}\implies\sin a=-\sqrt{\dfrac{144}{169}}=-\dfrac{12}{13}$

One final identity:

$\sin2a=2\sin a\cos a$

from which we get

$\sin2a=2\cdot\left(-\dfrac{12}{13}\right)\cdot\dfrac5{13}=-\dfrac{120}{169}$