Question 746467
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(A)\ =\ -\frac{8}{17}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(A)\ =\ \frac{64}{289}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \cos^2(A)\ =\ \frac{64}{289}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(A)\ =\ \frac{225}{289}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(A)\ =\ \pm\frac{15}{17}]


But since *[tex \LARGE \cos\varphi\ <\ 0] in quadrant III:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(A)\ =\ -\frac{15}{17}]


Use the same technique to find *[tex \LARGE \sin(B)]


Then use the formula for the cosine of a difference to calculate the desired value:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(\alpha\ -\ \beta)\ =\ \cos\alpha\cos\beta\ +\ \sin\alpha\sin\beta]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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