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Miscellaneous Exercise 3 (II) · Q112

Q.Prove the following: cos⁡36°=5+14\cos36°=\dfrac{\sqrt5+1}{4}

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Step 1: From the previous result, sin⁡18°=5−14\sin18°=\dfrac{\sqrt5-1}{4}.

Step 2: cos⁡36°=1−2sin⁡218°=1−2(5−14)2=1−2⋅6−2516=1−6−258\cos36°=1-2\sin^218°=1-2\left(\dfrac{\sqrt5-1}{4}\right)^2=1-2\cdot\dfrac{6-2\sqrt5}{16}=1-\dfrac{6-2\sqrt5}{8}. …

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