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Miscellaneous Exercise 3 (II) · Q113

Q.Prove the following: sin⁡36°=10−254\sin36°=\dfrac{\sqrt{10-2\sqrt5}}{4}

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Step 1: From the previous result, cos⁡36°=5+14\cos36°=\dfrac{\sqrt5+1}{4}.

Step 2: sin⁡236°=1−cos⁡236°=1−(5+14)2=1−6+2516=16−6−2516=10−2516\sin^236°=1-\cos^236°=1-\left(\dfrac{\sqrt5+1}{4}\right)^2=1-\dfrac{6+2\sqrt5}{16}=\dfrac{16-6-2\sqrt5}{16}=\dfrac{10-2\sqrt5}{16}. …

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