Question 516629
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First use the Pythagorean Identity to find *[tex \Large \cos\theta].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\varphi\ =\ 1\ -\ \sin^2\varphi]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\theta\ =\ 1\ -\ \frac{9}{25}\ =\ \frac{16}{25}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ \pm\frac{4}{5}]


But recall *[tex \Large 0\ <\ \theta\ <\ \frac{\pi}{2}], hence *[tex \Large \theta] is a 1st quadrant angle, therefore the cosine is positive.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ \frac{4}{5}]


Now use the half angle formula for cosine:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\frac{\theta}{2}\ =\ \pm\sqrt{\frac{1\ +\ \cos\theta}{2}}]


But recall *[tex \Large 0\ <\ \theta\ <\ \frac{\pi}{2}], hence *[tex \Large 0\ <\ \frac{\theta}{2}\ <\ \frac{\pi}{4}], is also a 1st quadrant angle, therefore the cosine is positive.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\frac{\theta}{2}\ =\ \sqrt{\frac{1\ +\ \frac{4}{5}}{2}}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\frac{\theta}{2}\ =\ \sqrt{\frac{9}{10}}\ =\ \frac{3\sqrt{10}}{10}]


After rationalizing the denominator.  One last thing:  pie is dessert, pi is the lower case Greek letter denoting the ratio between the circumference and the diameter of a circle.


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