$HSC Ext 1 Math Australia trigonometry question$

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$HSC Ext 1 Math Australia trigonometry question$

Prove that $\sin^2(\pi/18)+\sin^2(5\pi/18)+\sin^2(25\pi/18)=3/2$.

I believe the way im meant to solve this involves mainly the use of tried identities, and I have a strong feeling the triple angle identities have a part. Also this is a band 4 question worth 3 marks

Trigonometry

Jossunnys

7

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Kav10

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We have
\[\sin^2(\frac{\pi}{18})+\sin^2(\frac{5\pi}{18})+\sin^2(\frac{25\pi}{18})\]
\[=\frac{1-\cos (\frac{\pi}{9})}{2}+\frac{1-\cos (\frac{5\pi}{9})}{2}+\frac{1-\cos (\frac{25\pi}{9})}{2}\]
\[=\frac{3}{2}-\frac{1}{2}[ \cos (\frac{\pi}{9})+\cos (\frac{5\pi}{9})+\cos (\frac{7\pi}{9})]\]
\[=\frac{3}{2}-\frac{1}{2}[ \cos (\frac{\pi}{9})+2 \cos (\frac{\frac{7\pi}{9}+\frac{5\pi}{9}}{2}) \cos (\frac{\frac{7\pi}{9}-\frac{5\pi}{9}}{2})]\]
\[=\frac{3}{2}-\frac{1}{2}[ \cos (\frac{\pi}{9})+2 \cos (\frac{4\pi}{3}) \cos (\frac{\pi}{9})]\]
\[=\frac{3}{2}-\frac{1}{2}[ \cos (\frac{\pi}{9})+2 (-\cos (\frac{\pi}{3})) \cos (\frac{\pi}{9})]\]
\[=\frac{3}{2}-\frac{1}{2}[ \cos (\frac{\pi}{9})+2 (-\frac{1}{2}) \cos (\frac{\pi}{9})]\]
\[=\frac{3}{2}-\frac{1}{2}[ \cos (\frac{\pi}{9})- \cos (\frac{\pi}{9})]\]
\[=\frac{3}{2}.\]

Daniel90

432

Jossunnys

0

Can I ask what Identity you used in the first step to convert sin^2(π/18) to (1-cos(π/9))/2?
- Jossunnys
  
  +1
  
  Nvm I understand now, thanks

$HSC Ext 1 Math Australia trigonometry question$

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