# Thread: Puzzling Inverse Trig Function question #2

1. ## Puzzling Inverse Trig Function question #2

My question here is how they turned the answer into a fraction.

My process would go;

sin ( 2arccsc (13/5)) = sin (2 arcsin (5/13)) = sin (2 arcsin (0.3846153)) = sin (2 (22.619)) = sin (45.2397) = 0.71005917

How do you go about turning that into a fraction?

2. ## Re: Puzzling Inverse Trig Function question #2

let $\theta = \text{ arccsc}\left(\dfrac{13}{5}\right)$, assume $0<\theta<\dfrac{\pi}{2}$

$\sin(2\theta) = 2\sin{\theta} \cos{\theta}$

sketch a reference triangle in quad I (see attach)

$2 \cdot \dfrac{5}{13} \cdot \dfrac{12}{13} = \dfrac{120}{169}$

... have you studied identities yet?

3. ## Re: Puzzling Inverse Trig Function question #2

Originally Posted by blakesgrave

My question here is how they turned the answer into a fraction.
My process would go;
sin ( 2arccsc (13/5)) = sin (2 arcsin (5/13)) = sin (2 arcsin (0.3846153)) = sin (2 (22.619)) = sin (45.2397) = 0.71005917
How do you go about turning that into a fraction?
As I said in another reply: you are not prepared to do these questions.
You need to learn(memorize) these sum/difference and double angle formula.
$\sin(2A)=2\sin(a)\cos(A)~\&~\cos(2A)=\left\{ \begin{array}{l}\cos^2(A)-\sin^2(A)\\1-2\sin^2(A)\\2\cos^2(A)-1\end{array} \right.$