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Question Number 109923 by mathdave last updated on 26/Aug/20

$$find$$$$sin3$$$$insurdform$$

Answered by malwan last updated on 26/Aug/20

$$sin3°=sin(18-15)$$$$=sin18cos15-cos18sin15$$$$=\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{2+\sqrt{3}}}{2}\right)-\left(\frac{\sqrt{5+\sqrt{5}}}{2}\right)\left(\frac{\sqrt{2-\sqrt{3}}}{2}\right)$$$$=\frac{1}{48}\sqrt{6}(\sqrt{5}-1)(3+\sqrt{3})-$$$$\frac{1}{24}(\sqrt{3}(3-\sqrt{3})\left(\sqrt{5+\sqrt{5}}\right)$$

Commented by mathdave last updated on 26/Aug/20

$$correct$$

Commented by malwan last updated on 26/Aug/20

$$thankyousir$$

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