f67c8a8b-c2f4-4fc0-ae79-afe931b43ddf40d7c67b-1612-4c46-b645-18f0ba83cc206c89a17a-24f3-4809-bf53-6c9b648a61731.3.4.3<p>Calcular les raons trigonomètriques de l'angle de $$15$$ graus a partir de les raons dels angle de $$45$$ graus i $$30$$ graus.</p><p>Primer de tot, observem que $$15=45-30$$. Per tant, sabent les raons de $$45$$ graus i $$30$$ graus, a partir de les relacions de l'angle diferència, podrem obtenir fàcilment les de l'angle $$15^\circ$$. Calculem doncs: $$$\sin(45^\circ)=\dfrac{\sqrt{2}}{2} \ \ \ \ \sin(30^\circ)=\dfrac{1}{2}$$$ $$$\cos(45^\circ)=\dfrac{\sqrt{2}}{2} \ \ \ \ \cos(30^\circ)=\dfrac{\sqrt{3}}{2}$$$ $$$\tan(45^\circ)=1 \ \ \ \ \tan(30^\circ)=\dfrac{\sqrt{3}}{3}$$$</p> <p>A partir de les següents fórmules tenim: $$$\sin(15^\circ)=\sin(45^\circ-30^\circ)=\sin(45^\circ)\cdot \cos(30^\circ)-\cos(45^\circ)\cdot \sin(30^\circ)=$$$ $$$=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{2}\cdot\sqrt{3}}{4}-\dfrac{\sqrt{2}}{4}=\dfrac{\sqrt{2}}{4}(\sqrt{3}-1) \\ $$$</p> <p>$$$\cos(15^\circ)=\cos(45^\circ-30^\circ)=\cos(45^\circ)\cdot \cos(30^\circ)+\sin(45^\circ)\cdot \sin(30^\circ)=$$$ $$$=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{2}\cdot\sqrt{3}}{4}+\dfrac{\sqrt{2}}{4}=\dfrac{\sqrt{2}}{4}(\sqrt{3}+1) \\ $$$</p> <p>$$$\tan(15^\circ)=\tan(45^\circ-30^\circ)=\dfrac{\tan(45^\circ)-\tan(30^\circ)}{1+\tan(45^\circ)\cdot \tan(30^\circ)}=$$$ $$$\dfrac{1-\dfrac{\sqrt{3}}{3}}{1+\dfrac{\sqrt{3}}{3}}=\dfrac{\dfrac{3-\sqrt{3}}{3}}{\dfrac{3+\sqrt{3}}{3}}=\dfrac{3-\sqrt{3}}{3+\sqrt{3}}= \dfrac{3-\sqrt{3}}{3+\sqrt{3}}\cdot \dfrac{3-\sqrt{3}}{3-\sqrt{3}}=\dfrac{9+3-6\sqrt{3}}{9-3}=$$$ $$$=\dfrac{12-6\sqrt{3}}{6}=2-\sqrt{3}$$$</p><p>$$\sin(15^\circ)=\dfrac{\sqrt{2}}{4}(\sqrt{3}-1)$$</p> <p>$$\cos(15^\circ)=\dfrac{\sqrt{2}}{4}(\sqrt{3}+1)$$</p> <p>$$\tan(15^\circ)=2-\sqrt{3}$$</p>18/7/16 11:4418/7/16 11:44