Question

Oblicz [tex]\frac{3sin2\alpha -cos^{2}3\alpha }{tg4\alpha -\frac{1}{tg3\alpha } }[/tex]

jeżeli [tex]\alpha \alph[/tex]=15°

proszę o wytłumaczenie

Answer 1

Odpowiedź:

α = 15°   to   2α = 2*15° = 30°

3α = 45°°

4α = 60°

więc

w = [tex]\frac{3*sin 30^o - ( cos 45^o)^2}{tg 60^o - ctg 45^o} =\frac{3*0,5 - (\sqrt{2}/2)^2 }{\sqrt{3}- 1 } =[/tex]  [tex]\frac{1,5 - 0,5}{\sqrt{3} - 1} = \frac{1}{\sqrt{3}- 1} *\frac{\sqrt{3}+1 }{{\sqrt{3} +1} } =[/tex]

[tex]= \frac{\sqrt{3}+ 1 }{2}[/tex]

Szczegółowe wyjaśnienie:

[tex]\frac{1}{tg 3\alpha } = ctg 3\alpha[/tex]

( a - b) *(a + b) = a² - b²

Answer 2

[tex]\alpha = 15^{o}\\\\sin2\alpha = sin(2\cdot15^{o}) = sin30^{o} = \frac{1}{2}\\\\cos3\alpha = cos(3\cdot15^{o}) = cos45^{o} = \frac{\sqrt{2}}{2}\\\\tg3\alpha = tg(3\cdot15^{o}) = tg45^{o}=1\\\\tg4\alpha = tg(4\cdot15^{o}) = tg60^{o} = \sqrt{3}[/tex]

[tex]\frac{3sin2\alpha - cos^{2}3\alpha}{tg4\alpha -\frac{1}{tg3\alpha}}=\frac{3sin30^{o}-cos^{2}45^{o}}{tg60^{o}-\frac{1}{tg45^{o}}} = \frac{3\cdot\frac{1}{2}-(\frac{\sqrt{2}}{2})^{2}}{\sqrt{3}-\frac{1}{1}}=\frac{\frac{3}{2}-\frac{2}{4}}{\sqrt{3}-1} = \frac{\frac{6}{4}-\frac{2}{4}}{\sqrt{3}-1} = \frac{\frac{4}{4}}{\sqrt{3}-1} =\\\\= \frac{1}{\sqrt{3}-1}\cdot\frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{\sqrt{3}+1}{(\sqrt{3})^{2}-1^{2}} = \frac{\sqrt{3}+1}{3-1} = \boxed{\frac{\sqrt{3}+1}{2}}[/tex]