Question 1040805
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(a)\ =\ \frac{4}{5}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(a)\ =\ \frac{16}{25}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \1\ -\ \cos^2(a)\ =\ \frac{16}{25}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(a)\ =\ \frac{9}{25}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(a)\ =\ \pm\frac{3}{5}]


Similarly, you can show that *[tex \Large \sin(b)\ =\ \pm\frac{12}{13}]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(a)\ =\ \pm\frac{4/5}{3/5}\ =\ \pm\frac{4}{3}]


Similarly,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(b)\ =\ \pm\frac{12}{5}]


Tangent of the difference formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(a\ -\ b)\ =\ \frac{\tan(a)\ -\ \tan(b)}{1\ +\ \tan(a)\tan(b)}]


Case 1:  *[tex \LARGE \tan(a)\ >\ 0,\ \tan(b)\ >\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(a\ -\ b)\ =\ \frac{\frac{4}{3}\ -\ \frac{12}{5}}{1\ +\ \left(\frac{4}{3}\right)\left(\frac{12}{5}\right)}]


Just do the arithmetic.  Then plug in the numbers and do the arithmetic for the other three cases.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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