3)Przeksztalcamy do wspolnego mianownika:[tex]\frac{sin^{2}(a)+cos^{2}(a)}{cos(a)sin(a)}[/tex]
W liczniku jest jedynka trygonometryczna
[tex]\frac{1}{sin(a)cos(a)} = 2\\ sin(a)cos(a) = \frac{1}{2}[/tex]
5)
a)
[tex]sin\alpha (sin\alpha *\frac{sin\alpha }{cos\alpha }+cos\alpha ) = sin\alpha*\frac{sin^{2}a+cos^{2}a}{cos\alpha } = sin\alpha*\frac{1}{cos\alpha } =\\= \frac{sin\alpha }{cos\alpha } = tg\alpha[/tex]
b) Ze wzoru na roznice kwadratow dostajemy[tex]\frac{cos^{2}a-sin^{2}a}{cos^{2}a} = \frac{cos^{2}a}{cos^{2}a} - \frac{sin^{2}a}{cos^{2}a}[/tex]
Pierwszy ulamek wynosi 1, a drugi [tex]tg^{2}a[/tex]
Wowczas[tex]\frac{cos^{2}a}{cos^{2}a} - \frac{sin^{2}a}{cos^{2}a} = 1 - tg^{2}a[/tex]
c) [tex]\frac{1}{cos\alpha} - \frac{cos\alpha}{1+sin\alpha} = \frac{1+sin\alpha-cos^{2}\alpha }{cos\alpha(1+sin\alpha)} = \frac{sin\alpha + 1 -cos^{2}\alpha }{cos\alpha(1+sin\alpha)} = \frac{sin\alpha +sin^{2}\alpha }{cos\alpha(1+sin\alpha)} = \frac{sin\alpha (1+sin\alpha )}{cos\alpha(1+sin\alpha)}[/tex]
[tex]\frac{sin\alpha (1+sin\alpha )}{cos\alpha(1+sin\alpha)} = \frac{sin\alpha }{cos\alpha } =tg\alpha[/tex]
6) Tu nalezy skorzystac ze wzoru [tex]sin2\alpha = 2sin\alpha cos\alpha[/tex]
[tex]sin10cos10cos20cos40 = \frac{1}{2}sin20cos20cos40 = \frac{1}{4}sin40cos40 = \frac{1}{8}sin80[/tex]
7) [tex]\frac{cos\alpha +sina}{cos\alpha } = \frac{cos\alpha }{cos\alpha } + \frac{sin\alpha }{cos\alpha } = 1 + tg \alpha = 1 +\frac{1}{ctg\alpha } = 1 + \frac{1}{2}=\frac{3}{2}[/tex]
8) [tex]cos40 = cos(90-50) = sin50\\\alpha =50[/tex]
9)[tex]\frac{3}{2} + cos\alpha = 2\\cos\alpha = \frac{1}{2}\\ \alpha = 60[/tex]
10)[tex](sin\alpha -cos\alpha )^{2} = sin^{2}\alpha -2sin\alpha cos\alpha +cos^{2}\alpha = sin^{2}\alpha+cos^{2}\alpha -2sin\alpha cos\alpha=\\= 1 - 2*\frac{1}{4} = \frac{1}{2}[/tex]
1) [tex]\frac{sin\alpha -cos\alpha }{sin\alpha + cos\alpha}[/tex]
Mnozymy licznik oraz mianownik na ulamek [tex]\frac{1}{cos\alpha }[/tex]
[tex]\frac{sin\alpha\frac{1}{cos\alpha } -1 }{sin\alpha\frac{1}{cos\alpha }+ 1} = \frac{tg\alpha -1}{tg\alpha +1} = \frac{1}{3}[/tex]
2) [tex]4sin^2\alpha -3cos^2{\alpha } = 3\\[/tex]
Dzielimy obu czesci rownania przez [tex]cos^{2}\alpha[/tex]
[tex]4tg^2\alpha -3 = 3*\frac{1}{cos^{2}\alpha } \\4tg^2\alpha -3 = 3 * \frac{sin^2\alpha +cos^2\alpha }{cos^{2}\alpha } \\4tg^2\alpha -3 = 3(tg^2\alpha +1)\\4tg^2\alpha -3 = 3tg^2a+3\\tg^2a = 6\\tg\alpha = \sqrt{6}[/tex]
Mysle ze tego wystarczy. Daj mi znac czy wszystko jest zrozumiale
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3)
Przeksztalcamy do wspolnego mianownika:
[tex]\frac{sin^{2}(a)+cos^{2}(a)}{cos(a)sin(a)}[/tex]
W liczniku jest jedynka trygonometryczna
[tex]\frac{1}{sin(a)cos(a)} = 2\\ sin(a)cos(a) = \frac{1}{2}[/tex]
5)
a)
[tex]sin\alpha (sin\alpha *\frac{sin\alpha }{cos\alpha }+cos\alpha ) = sin\alpha*\frac{sin^{2}a+cos^{2}a}{cos\alpha } = sin\alpha*\frac{1}{cos\alpha } =\\= \frac{sin\alpha }{cos\alpha } = tg\alpha[/tex]
b) Ze wzoru na roznice kwadratow dostajemy
[tex]\frac{cos^{2}a-sin^{2}a}{cos^{2}a} = \frac{cos^{2}a}{cos^{2}a} - \frac{sin^{2}a}{cos^{2}a}[/tex]
Pierwszy ulamek wynosi 1, a drugi [tex]tg^{2}a[/tex]
Wowczas
[tex]\frac{cos^{2}a}{cos^{2}a} - \frac{sin^{2}a}{cos^{2}a} = 1 - tg^{2}a[/tex]
c) [tex]\frac{1}{cos\alpha} - \frac{cos\alpha}{1+sin\alpha} = \frac{1+sin\alpha-cos^{2}\alpha }{cos\alpha(1+sin\alpha)} = \frac{sin\alpha + 1 -cos^{2}\alpha }{cos\alpha(1+sin\alpha)} = \frac{sin\alpha +sin^{2}\alpha }{cos\alpha(1+sin\alpha)} = \frac{sin\alpha (1+sin\alpha )}{cos\alpha(1+sin\alpha)}[/tex]
[tex]\frac{sin\alpha (1+sin\alpha )}{cos\alpha(1+sin\alpha)} = \frac{sin\alpha }{cos\alpha } =tg\alpha[/tex]
6) Tu nalezy skorzystac ze wzoru [tex]sin2\alpha = 2sin\alpha cos\alpha[/tex]
[tex]sin10cos10cos20cos40 = \frac{1}{2}sin20cos20cos40 = \frac{1}{4}sin40cos40 = \frac{1}{8}sin80[/tex]
7) [tex]\frac{cos\alpha +sina}{cos\alpha } = \frac{cos\alpha }{cos\alpha } + \frac{sin\alpha }{cos\alpha } = 1 + tg \alpha = 1 +\frac{1}{ctg\alpha } = 1 + \frac{1}{2}=\frac{3}{2}[/tex]
8) [tex]cos40 = cos(90-50) = sin50\\\alpha =50[/tex]
9)[tex]\frac{3}{2} + cos\alpha = 2\\cos\alpha = \frac{1}{2}\\ \alpha = 60[/tex]
10)[tex](sin\alpha -cos\alpha )^{2} = sin^{2}\alpha -2sin\alpha cos\alpha +cos^{2}\alpha = sin^{2}\alpha+cos^{2}\alpha -2sin\alpha cos\alpha=\\= 1 - 2*\frac{1}{4} = \frac{1}{2}[/tex]
1) [tex]\frac{sin\alpha -cos\alpha }{sin\alpha + cos\alpha}[/tex]
Mnozymy licznik oraz mianownik na ulamek [tex]\frac{1}{cos\alpha }[/tex]
[tex]\frac{sin\alpha\frac{1}{cos\alpha } -1 }{sin\alpha\frac{1}{cos\alpha }+ 1} = \frac{tg\alpha -1}{tg\alpha +1} = \frac{1}{3}[/tex]
2) [tex]4sin^2\alpha -3cos^2{\alpha } = 3\\[/tex]
Dzielimy obu czesci rownania przez [tex]cos^{2}\alpha[/tex]
[tex]4tg^2\alpha -3 = 3*\frac{1}{cos^{2}\alpha } \\4tg^2\alpha -3 = 3 * \frac{sin^2\alpha +cos^2\alpha }{cos^{2}\alpha } \\4tg^2\alpha -3 = 3(tg^2\alpha +1)\\4tg^2\alpha -3 = 3tg^2a+3\\tg^2a = 6\\tg\alpha = \sqrt{6}[/tex]
Mysle ze tego wystarczy. Daj mi znac czy wszystko jest zrozumiale