Posted by **sh** on Friday, March 5, 2010 at 3:20am.

Find the value of tan2x, (pi/2)<x<pi, given secx=-5/4.

So cosx=-4/5? I've no clue what to do next.

- Calculus -
**Reiny**, Friday, March 5, 2010 at 7:42am
According to the given domain, the angle must be in quadrant II,

We are given cosx = - 4/5.

remember that cosine is adjacent/hypotenuse, so construct a right angles triangle with base of 4 and hypotenuse of 5, you should recognize the 3-4-5 rightangled triangle so the height (or opposite) is 3 and

sinx = 3/5

remember that tangent = sine/cosine

and tanx = (3/5)/(-4/5) = - 3/4

have you come across the identiy

tan 2x = 2tanx/(1 - tan^2 x) ?

tan2x = 2(-3/4)/(1 - 9/16)

= (-3/2)/(7/16)

= -24/7

- Calculus -
**sh**, Friday, March 5, 2010 at 9:55am
Thanks!

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