# Thread: csc (A - B)

1. ## csc (A - B)

If A = 45 degrees and B = arcsin(3/5), find the exact value of csc (A - B).

2. Originally Posted by magentarita
If A = 45 degrees and B = arcsin(3/5), find the exact value of csc (A - B).
Hints:

$\csc{X} = \frac{1}{\sin{X}}$

$\sin{(\alpha - \beta)} = \sin{\alpha}\cos{\beta}- \cos{\alpha}\sin{\beta}$.

3. ## Yes but...

Originally Posted by Prove It
Hints:

$\csc{X} = \frac{1}{\sin{X}}$

$\sin{(\alpha - \beta)} = \sin{\alpha}\cos{\beta}- \cos{\alpha}\sin{\beta}$.
My problem with this question is arcsin(3/5). In the given formula, A = 45 and B = arcsin(3/5). I can't go on without knowing the value of arcsin(3/5).

4. Originally Posted by magentarita
My problem with this question is arcsin(3/5). In the given formula, A = 45 and B = arcsin(3/5). I can't go on without knowing the value of arcsin(3/5).
Yes you can.

Remember that $\sin{(\arcsin{\theta})} = \theta$ and $\cos{\theta} = \sin{(90^\circ - \theta)}$.

5. ## ok

I will play with this question and get back to you later.

6. Originally Posted by magentarita
I will play with this question and get back to you later.
Note that $B = \arcsin \left( \frac{3}{5}\right) \Rightarrow \sin B = \frac{3}{5}$.

And $\sin B = \frac{3}{5} \Rightarrow \cos B = \frac{4}{5}$ (the postive value is taken because the range of the arcsin function is angles in the 4th and 1st quadrants and cos is postive in these two quadrants).