csc (A - B)

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May 5th 2009, 10:21 PM
magentarita

csc (A - B)

If A = 45 degrees and B = arcsin(3/5), find the exact value of csc (A - B).
May 5th 2009, 10:33 PM
Prove It

Quote:

Originally Posted by magentarita

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

Hints:

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

$\sin{(\alpha - \beta)} = \sin{\alpha}\cos{\beta}- \cos{\alpha}\sin{\beta}$ .
May 6th 2009, 06:20 AM
magentarita

Yes but...

Quote:

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).
May 6th 2009, 03:42 PM
Prove It

Quote:

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 $\cos{\theta} = \sin{(90^\circ - \theta)}$ .
May 7th 2009, 07:11 AM
magentarita

ok

I will play with this question and get back to you later.
May 7th 2009, 07:15 AM
mr fantastic

Quote:

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).

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