Trigonometry Challenge 1

Geometry Level 3

csc ⁡ π 18 − 3 sec ⁡ π 18 = ? \csc {\frac{\pi}{18}}-\sqrt{3}\sec {\frac{\pi}{18}}=\boxed{?}

Clarifications:

  1. csc ⁡ ( ⋅ ) \csc(\cdot) denotes the cosecant function.
  2. All angles are in radians.


The answer is 4.

Noel Lo by Noel Lo , 24, Singapore

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2 solutions

Noel Lo
Jul 18, 2017

csc ⁡ π 18 − 3 sec ⁡ π 18 = 1 sin ⁡ π 18 − 3 cos ⁡ π 18 = cos ⁡ π 18 − 3 sin ⁡ π 18 sin ⁡ π 18 cos ⁡ π 18 = 2 ( 1 2 cos ⁡ π 18 − 3 2 sin ⁡ π 18 ) sin ⁡ π 18 cos ⁡ π 18 = 2 ( sin ⁡ π 6 cos ⁡ π 18 − cos ⁡ π 6 sin ⁡ π 18 ) sin ⁡ π 18 cos ⁡ π 18 \csc {\frac{\pi}{18}}-\sqrt{3}\sec {\frac{\pi}{18}}=\frac{1}{\sin {\frac{\pi}{18}}}-\frac{\sqrt{3}}{\cos {\frac{\pi}{18}}}=\frac{\cos {\frac{\pi}{18}}-\sqrt{3}\sin {\frac{\pi}{18}}}{\sin {\frac{\pi}{18}}\cos {\frac{\pi}{18}}}=\frac{2(\frac{1}{2}\cos {\frac{\pi}{18}}-\frac{\sqrt{3}}{2}\sin {\frac{\pi}{18}})}{\sin {\frac{\pi}{18}}\cos {\frac{\pi}{18}}}= \frac{2(\sin{\frac{\pi}{6}}\cos {\frac{\pi}{18}}-\cos{\frac{\pi}{6}}\sin {\frac{\pi}{18}})}{\sin {\frac{\pi}{18}}\cos {\frac{\pi}{18}}}

= 2 sin ⁡ ( π 6 − π 18 ) sin ⁡ π 18 cos ⁡ π 18 = 2 sin ⁡ π 9 sin ⁡ π 18 cos ⁡ π 18 = 2 ( 2 sin ⁡ π 9 ) 2 sin ⁡ π 18 cos ⁡ π 18 = ( 2 × 2 ) sin ⁡ π 9 sin ⁡ ( 2 × π 18 ) = 4 sin ⁡ π 9 sin ⁡ π 9 = 4 =\frac{2\sin{(\frac{\pi}{6}-\frac{\pi}{18}})}{\sin {\frac{\pi}{18}}\cos {\frac{\pi}{18}}}=\frac{2\sin{\frac{\pi}{9}}}{\sin {\frac{\pi}{18}}\cos {\frac{\pi}{18}}}=\frac{2(2\sin{\frac{\pi}{9}})}{2\sin {\frac{\pi}{18}}\cos {\frac{\pi}{18}}}=\frac{(2 \times 2)\sin{\frac{\pi}{9}}}{\sin {(2\times \frac{\pi}{18})}}=\frac{4\sin{\frac{\pi}{9}}}{\sin {\frac{\pi}{9}}}=\boxed{4}

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Chew-Seong Cheong
Jul 28, 2017

csc ⁡ π 18 − 3 sec ⁡ π 18 = 1 sin ⁡ π 18 − 3 cos ⁡ π 18 = cos ⁡ π 18 − 3 sin ⁡ π 18 sin ⁡ π 18 cos ⁡ π 18 = 2 ( 1 2 cos ⁡ π 18 − 3 2 sin ⁡ π 18 ) 1 2 sin ⁡ π 9 = 4 sin ⁡ ( π 6 − π 18 ) sin ⁡ π 9 = 4 sin ⁡ π 9 sin ⁡ π 9 = 4 \begin{aligned} \csc \frac \pi{18} - \sqrt 3 \sec \frac \pi{18} & = \frac 1{\sin \frac \pi{18}} - \frac {\sqrt 3}{\cos \frac \pi{18}} \\ & = \frac {\cos \frac \pi{18} - \sqrt 3 \sin \frac \pi{18}}{\sin \frac \pi{18}\cos \frac \pi{18}} \\ & = \frac {2\left(\frac 12 \cos \frac \pi{18} - \frac {\sqrt 3}2 \sin \frac \pi{18}\right)}{\frac 12 \sin \frac \pi 9} \\ & = \frac {4 \sin \left(\frac \pi 6 - \frac \pi{18}\right)}{\sin \frac \pi 9} \\ & = \frac {4 \sin \frac \pi 9}{\sin \frac \pi 9} \\ & = \boxed{4} \end{aligned}

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