Question 231699
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 105\ =\ 60\ +\ 45]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos{(45)}\ =\ \frac{\sqrt{2}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos{(60)}\ =\ \frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin{(45)}\ =\ \frac{\sqrt{2}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin{(60)}\ =\ \frac{\sqrt{3}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos{(45\ +\ 60)}\ =\ \cos{(45)}\cos{(60)}\ -\ \sin{(45)}\sin{(60)}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos{(105)}\ =\ \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)\ -\ \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)]


Simplification of this expression is left as an exercise for the student.


Sanity check:  Notice that the left hand term is smaller than the right hand term.  That means that cos(105) < 0, which makes sense for a Quadrant II angle.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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