Maths Olympiad Prep

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, 2008

Geometry Difficulty 5.7 AIME, harder Prove it Slovenia

Prove that for all real numbers x∈(−3π2,π2)x \in \left(-\frac{3\pi}{2}, \frac{\pi}{2}\right) the equality
21−sin⁡x=tan⁡2(x2+π4)+1 \frac{2}{1 - \sin x} = \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) + 1
holds.

Solution

From the addition formula for tangents we get
tan⁡2(x2+π4)=(tan⁡x2+tan⁡π41−tan⁡x2tan⁡π4)2. \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \left( \frac{\tan \frac{x}{2} + \tan \frac{\pi}{4}}{1 - \tan \frac{x}{2} \tan \frac{\pi}{4}} \right)^2 .
Using the fact that tan⁡π4=1\tan \frac{\pi}{4} = 1 and expressing tan⁡x2\tan \frac{x}{2} in terms of sines and cosines, we see that
tan⁡2(x2+π4)=(sin⁡x2cos⁡x2+11−sin⁡x2cos⁡x2)2=(cos⁡x2+sin⁡x2cos⁡x2−sin⁡x2)2. \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \left( \frac{\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} + 1}{1 - \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}} \right)^2 = \left( \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \right)^2 .
We now square both sides of the equation to get
tan⁡2(x2+π4)=cos⁡2x2+2cos⁡x2sin⁡x2+sin⁡2x2cos⁡2x2−2cos⁡x2sin⁡x2+sin⁡2x2 \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \frac{\cos^2 \frac{x}{2} + 2 \cos \frac{x}{2} \sin \frac{x}{2} + \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} - 2 \cos \frac{x}{2} \sin \frac{x}{2} + \sin^2 \frac{x}{2}}
and use the relations cos⁡2x2+sin⁡2x2=1\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} = 1 and 2cos⁡x2sin⁡x2=sin⁡x2 \cos \frac{x}{2} \sin \frac{x}{2} = \sin x to show that
tan⁡2(x2+π4)=1+sin⁡x1−sin⁡x=21−sin⁡x−1. \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \frac{1 + \sin x}{1 - \sin x} = \frac{2}{1 - \sin x} - 1.
Thus, the equality holds.

Solution 2:
Express the tangent in terms of sine and cosine
tan⁡2(x2+π4)=(sin⁡(x2+π4)cos⁡(x2+π4))2. \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \left( \frac{\sin \left( \frac{x}{2} + \frac{\pi}{4} \right)}{\cos \left( \frac{x}{2} + \frac{\pi}{4} \right)} \right)^2 .
Now, use the addition formulas for sine and cosine to show that
tan⁡2(x2+π4)=(sin⁡x2cos⁡π4+cos⁡x2sin⁡π4cos⁡x2cos⁡π4−sin⁡x2sin⁡π4)2. \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \left( \frac{\sin \frac{x}{2} \cos \frac{\pi}{4} + \cos \frac{x}{2} \sin \frac{\pi}{4}}{\cos \frac{x}{2} \cos \frac{\pi}{4} - \sin \frac{x}{2} \sin \frac{\pi}{4}} \right)^2 .
We know that sin⁡π4=22\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} and cos⁡π4=22\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, so the expression is equal to
tan⁡2(x2+π4)=(22sin⁡x2+22cos⁡x222cos⁡x2−22sin⁡x2)2. \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \left( \frac{\frac{\sqrt{2}}{2} \sin \frac{x}{2} + \frac{\sqrt{2}}{2} \cos \frac{x}{2}}{\frac{\sqrt{2}}{2} \cos \frac{x}{2} - \frac{\sqrt{2}}{2} \sin \frac{x}{2}} \right)^2 .
Squaring both sides we get
tan⁡2(x2+π4)=cos⁡2x2+2cos⁡x2sin⁡x2+sin⁡2x2cos⁡2x2−2cos⁡x2sin⁡x2+sin⁡2x2 \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \frac{\cos^2 \frac{x}{2} + 2 \cos \frac{x}{2} \sin \frac{x}{2} + \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} - 2 \cos \frac{x}{2} \sin \frac{x}{2} + \sin^2 \frac{x}{2}}
and the equalities cos⁡2x2+sin⁡2x2=1\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} = 1 and 2cos⁡x2sin⁡x2=sin⁡x2 \cos \frac{x}{2} \sin \frac{x}{2} = \sin x imply
tan⁡2(x2+π4)=1+sin⁡x1−sin⁡x=21−sin⁡x−1. \tan^2 \left( \frac{x}{2} + \frac{\pi}{4} \right) = \frac{1 + \sin x}{1 - \sin x} = \frac{2}{1 - \sin x} - 1.

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