Trigonometry! #7

Geometry Level 4

Let α , β \alpha, \beta be such that π < α − β < 3 π \pi < \alpha - \beta < 3 \pi .

If sin ⁡ α + sin ⁡ β = − 21 65 \sin \alpha + \sin \beta = -\frac {21}{65} and cos ⁡ α + cos ⁡ β = − 27 65 \cos \alpha + \cos \beta = -\frac {27}{65} , then cos ⁡ ( α − β 2 ) \cos (\frac {\alpha - \beta}{2}) is

The calculations are pretty complex, so I advise you to go open a calculator.

This problem is part of the set Trigonometry .

− 6 65 -\frac {6}{\sqrt{65}} 6 65 \frac {6}{\sqrt {65}} 3 130 \frac {3}{\sqrt{130}} − 3 130 -\frac {3}{\sqrt{130}}

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Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Omkar Kulkarni
Jan 28, 2015

When we simplify sin ⁡ α + sin ⁡ β = − 21 65 \sin \alpha + \sin \beta = -\frac {21}{65} and take squares on both sides, we get

sin ⁡ 2 ( α + β 2 ) cos ⁡ 2 ( α − β 2 ) = 441 16900 \sin^{2}\left(\frac {\alpha + \beta}{2} \right)\cos^{2}\left(\frac {\alpha-\beta}{2}\right)=\frac {441}{16900}

Similarly, cos ⁡ α + cos ⁡ β = − 27 65 \cos \alpha + \cos \beta = -\frac {27}{65} becomes

cos ⁡ 2 ( α + β 2 ) cos ⁡ 2 ( α − β 2 ) = 729 16900 \cos^{2}\left(\frac {\alpha + \beta}{2} \right)\cos^{2}\left(\frac {\alpha-\beta}{2}\right)=\frac {729}{16900}

Adding both these equations, we get cos ⁡ ( α − β 2 ) = ± 3 130 \cos \left (\frac {\alpha-\beta}{2}\right)=±\frac{3}{\sqrt{130}}

As π < α − β < 3 π \pi<\alpha-\beta<3\pi , π 2 < α − β 2 < 3 π 2 \frac {\pi}{2}<\frac{\alpha-\beta}{2}<\frac{3\pi}{2} , and hence the answer is − 3 130 \boxed{-\frac {3}{\sqrt{130}}}

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