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Szczegółowe wyjaśnienie:
[tex]\frac{3sin2\alpha -cos^23\alpha }{tg4\alpha -\frac{1}{tg3\alpha } }[/tex] gdzie α=15°
[tex]\frac{3sin(2*15')-cos^2(3*15')}{tg(4*15')-\frac{1}{tg(3*15')} } =\frac{3sin45'-(cos45')^2}{tg60'-\frac{1}{tg45'} } =\frac{3*\frac{\sqrt{2} }{2} -(\frac{\sqrt{2} }{2})^2}{\sqrt{3} -\frac{1}{1} } =\frac{\frac{3\sqrt{2} }{2}-\frac{2}{4} }{\sqrt{3} -1} =[/tex]
[tex]=\frac{\frac{3\sqrt{2} }{2} -\frac{1}{2} }{\sqrt{3} -1} *\frac{\sqrt{3} +1}{\sqrt{3} +1} =\frac{\frac{1}{2} (3\sqrt{2} -1)(\sqrt{3} +1)}{3-1} =\frac{1}{2} *\frac{3\sqrt{6}+3\sqrt{2}-\sqrt{3} -1 }{2} =\frac{1}{4} (3\sqrt{6}+3\sqrt{2}-\sqrt{3} -1 )[/tex]
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Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]\frac{3sin2\alpha -cos^23\alpha }{tg4\alpha -\frac{1}{tg3\alpha } }[/tex] gdzie α=15°
[tex]\frac{3sin(2*15')-cos^2(3*15')}{tg(4*15')-\frac{1}{tg(3*15')} } =\frac{3sin45'-(cos45')^2}{tg60'-\frac{1}{tg45'} } =\frac{3*\frac{\sqrt{2} }{2} -(\frac{\sqrt{2} }{2})^2}{\sqrt{3} -\frac{1}{1} } =\frac{\frac{3\sqrt{2} }{2}-\frac{2}{4} }{\sqrt{3} -1} =[/tex]
[tex]=\frac{\frac{3\sqrt{2} }{2} -\frac{1}{2} }{\sqrt{3} -1} *\frac{\sqrt{3} +1}{\sqrt{3} +1} =\frac{\frac{1}{2} (3\sqrt{2} -1)(\sqrt{3} +1)}{3-1} =\frac{1}{2} *\frac{3\sqrt{6}+3\sqrt{2}-\sqrt{3} -1 }{2} =\frac{1}{4} (3\sqrt{6}+3\sqrt{2}-\sqrt{3} -1 )[/tex]