Question 1117332
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \tan(A)\ =\ \frac{3}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{\sin(A)}{\cos(A)}\ =\ \frac{3}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4\sin(A)\ =\ 3\cos(A)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  16\sin^2(A)\ =\ 9\cos^2(A)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  16\sin^2(A)\ =\ 9\(1\ -\ \sin^2(A)\)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  25\sin^2(A)\ =\ 9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin(A)\ =\ \pm\sqrt{\frac{9}{25}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  A\ \in\ \text{QI}\ \Right\ \sin(A)\ >\ 0\ \Right\ \sin(A)\ =\ \frac{3}{5}]


Use a similar argument to demonstrate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(A)\ =\ \frac{4}{5}]


And once more:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \cos(B)\ =\ -\frac{4}{5}]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin^2(B)\ =\ 1\ -\ \frac{16}{25}\ =\ \frac{9}{25}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin(B)\ =\ \pm\sqrt{\frac{9}{25}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  B\ \in\ \text{QIII}\ \Right\ \sin(B)\ <\ 0\ \Right\ \sin(B)\ =\ -\frac{3}{5}]


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \sin(A\ +\ B)\ =\ \sin(A)\cos(B)\ +\ \sin(B)\cos(A)]


Plug in the numbers and do the arithmetic.  Mind your signs



John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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