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Exercise 3.3 · Q6

Q.Show that sin⁡2π18+sin⁡2π9+sin⁡27π18+sin⁡24π9=2.\sin^2\dfrac{\pi}{18}+\sin^2\dfrac{\pi}{9}+\sin^2\dfrac{7\pi}{18}+\sin^2\dfrac{4\pi}{9}=2.

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Converting the third and fourth terms into cosines of the first and second angles (via the complementary-angle identity sin⁡(π2−θ)=cos⁡θ\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta) turns the sum into (sin⁡2A+cos⁡2A)+(sin⁡2B+cos⁡2B)=1+1=2(\sin^2A+\cos^2A)+(\sin^2B+\cos^2B)=1+1=2.

Step 1. Convert all four angles to degrees to see the complementary pairing clearly. π18=10∘, π9=20∘, 7π18=70∘, 4π9=80∘\dfrac{\pi}{18}=10^\circ,\ \dfrac{\pi}{9}=20^\circ,\ \dfrac{7\pi}{18}=70^\circ,\ \dfrac{4\pi}{9}=80^\circ.

Step 2. Spot the complementary pairs. 70∘=90∘−20∘70^\circ=90^\circ-20^\circ and 80∘=90∘−10∘80^\circ=90^\circ-10^\circ — so 70∘70^\circ is complementary to 20∘20^\circ, and 80∘80^\circ is complementary to 10∘10^\circ.

Step 3. Rewrite the third and fourth terms using sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ-\theta)=\cos\theta.

sin⁡70∘=sin⁡(90∘−20∘)=cos⁡20∘,sin⁡80∘=sin⁡(90∘−10∘)=cos⁡10∘.\sin70^\circ=\sin(90^\circ-20^\circ)=\cos20^\circ,\qquad \sin80^\circ=\sin(90^\circ-10^\circ)=\cos10^\circ.

So sin⁡27π18=cos⁡2π9\sin^2\dfrac{7\pi}{18}=\cos^2\dfrac{\pi}{9} and sin⁡24π9=cos⁡2π18\sin^2\dfrac{4\pi}{9}=\cos^2\dfrac{\pi}{18}.

Step 4. Substitute back into the sum and regroup. …

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