sin(2sec^-1(-5/2))

Question

sin(2sec^-1(-5/2))

tkhunny · Answer

You must draw a right triangle and label an acute angle $\alpha$.

Observe that $\sec(\alpha) = \dfrac{Hypotenuse}{Adjacent}$

Label the length fo the hypotenuse '5' and the length of the adjacent leg '2'

Use the Pythagorean theorem to calculate the length of the Opposite leg.

Now what?

anonymous · Answer

For the opposite side i got sqrt(21) then i plugged it into sin(x)=sqrt(21)/5 but i don't know what to do with the 2 in (2sec^-1(-5/2))

tkhunny · Answer

Got a little jumpy there, didn't you?

First, you have to convince yourself that the inverse secant function produces an ANGLE.  You put in a number, the secant, and you get out an angle that has the number as its secant.  $Angle = \sec^{-1}(Number)$.  Do you believe?

tkhunny · Answer

After that convincing, you can use the tried and true double angle identity.

tkhunny · Answer

$\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)$

tkhunny · Answer

After that, there is one more thing.  The original secant was negative?  In what quadrant does that put the inverse secant?  If you then double that angle, what quadrant are you in?  You will need this information to get the sign right.

Good luck.  Having connection problems.