My Fifth Problem

Geometry Level pending

Find:

s i n 2 ( a r c c o s 2 ) sin^{2}(arccos\sqrt{2})


Treat these as complex valued functions.


The answer is -1.

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1 solution

Daniel Ellesar
Jan 26, 2015

theta is the angle.

Draw a triangle, where since c o s θ = 2 cos \theta =\sqrt{2} , the hypotenuse is 1 and the adjacent is 2 \sqrt{2} .

Using Pythagoras we find that the opposite is i, so s i n θ sin \theta is i 1 = i \frac{i}{1} = i

s i n 2 θ sin^{2} \theta is -1

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