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Prove without using trigonometric table

sin² 5°+ sin² 10°+....+ sin² 85°+ sin² 90° = 9 \frac{1}{2}

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sin² 5°+ sin² 10°+....+ sin² 85°+ sin² 90° = 9 \frac{1}{2}

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so we have pairs: 5,85; 10,80, .... 40,50; and only one sin² 45 and one sin² 90

So the