Stuck on this problem, which identity should I be using?

[SCAZ]

If tan(A)=(-15/8) and A is in quadrant 4, find sin(2A). Do I use sin(2A)=2sin(A)cos(A) or a half angle identity like tan(A/2)? I've tried both and I'm lost.

 

[MarkFL]

Hello, and welcome to FMH!

Let's write:

[MATH]\sin(2A)=2\sin(A)\cos(A)=\frac{2\tan(A)}{\sec^2(A)}[/MATH]
Can we express the denominator as a function of $\tan(A)$?

 

[SCAZ]

I'm unsure of how to type my work into text here. I used the Pythagorean identity to find sec=(17/8). I plugged that into the equation and ended up with the final answer of (-30/17). Correct??

 

[Deleted member 4993]

If tan(A)=(-15/8) and A is in quadrant 4, find sin(2A). Do I use sin(2A)=2sin(A)cos(A) or a half angle identity like tan(A/2)? I've tried both and I'm lost.

Note that if A is in 4th quadrant then 2A is in 7 th (3 rd) or 8th (4 th) quadrant. In either case Sine will be <0.

Note if tan(A) = x then

$\displaystyle cos^2(A) = \dfrac{1}{1+x^2}$

You can calculate Cos(A) and Sin(A) from that hence calculate Sin(2A)

 

[Harry_the_cat]

I'm unsure of how to type my work into text here. I used the Pythagorean identity to find sec=(17/8). I plugged that into the equation and ended up with the final answer of (-30/17). Correct??

-30/17 can't be sin(2A) since it is not between -1 and 1 (incl).

Yes. sec(A) = 17/8, so cos(A) = 8/17 (positive since A is in the 4th quadrant. Now find sin (A) remembering that A is in the 4th quadrant, and then use the double angle identity to find sin(2A).

 

[MarkFL]

I'm unsure of how to type my work into text here. I used the Pythagorean identity to find sec=(17/8). I plugged that into the equation and ended up with the final answer of (-30/17). Correct??

What I was hoping you would see is that:

[MATH]\sec^2(A)=1+\tan^2(A)[/MATH]
and write:

[MATH]\sin(2A)=\frac{2\tan(A)}{\sec^2(A)}=\frac{2\tan(A)}{1+\tan^2(A)}=\frac{2\left(-\dfrac{15}{8}\right)}{1+\left(-\dfrac{15}{8}\right)^2}=-\frac{240}{289}[/MATH]
This is what you would have gotten if you had squared the value you found for $\sec(A)$.

 

[Steven G]

Draw a right triangle that agree with tan(A) = -15/8
Label the 3rd side (label ALL 3 sides). Then use sin(2A) = 2sin(A)cos(A)

 

[pka]

If tan(A)=(-15/8) and A is in quadrant 4, find sin(2A). Do I use sin(2A)=2sin(A)cos(A) or a half angle identity like tan(A/2)? I've tried both and I'm lost.

If we look at $\displaystyle \arctan\left(\frac{-15}{8}\right)$ the result is not nice. SEE HERE
So let's agree that $\displaystyle A=\arctan\left(\frac{-15}{8}\right)$
So that $\displaystyle \sin(2A)=2\sin(A)\cos(A)$ SEE HERE

 

[Dr.Peterson]

I'm unsure of how to type my work into text here. I used the Pythagorean identity to find sec=(17/8). I plugged that into the equation and ended up with the final answer of (-30/17). Correct??

It looks to me like you found 2 sin(A) instead of sin(2A). Or maybe you used MarkFL's formula and forgot to square the secant.

Jomo's way works very nicely for this problem. It's just what I did, and avoids complicated formulas.