csc (A - B)

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

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Originally Posted by magentarita

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

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Hints:

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


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

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Originally Posted by Prove It

Hints:

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


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

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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).

Quote:

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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).

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Yes you can.

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

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

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Originally Posted by magentarita

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

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Note that $\displaystyle B = \arcsin \left( \frac{3}{5}\right) \Rightarrow \sin B = \frac{3}{5}$.

And $\displaystyle \sin B = \frac{3}{5} \Rightarrow \cos B = \frac{4}{5}$ (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).