Dowodzenie tożsamości trygonometrycznych

[Erdbeeree]

1.wykaż tożsamość: \(sin^{4} \alpha -cos^{4} \alpha=2sin^{2} \alpha-1\)
b)\(\frac{sin5\alpha+sin3\alpha}{cos5\alpha+cos3\alpha}=tg4\alpha\)

c)\(tg\alpha*sin2\alpha=2sin^{2}\alpha\)

d)\(tg(45^{ \circ }+\alpha)*tg(45^{ \circ }-\alpha)=1\)

2.Udowodnij,że:
a)\(\frac{cos\alpha-cos \beta }{sin\alpha+sin\beta}=tg \frac{\beta-\alpha}{2}\)

b)\(\frac{cos(\alpha- \beta)+cos(\alpha+\beta)}{sin(\alpha-\beta)+sin(\alpha+\beta)}=ctg\alpha\)

c)\(\frac{tg^{2}(45^{ \circ }+\alpha)-1}{tg^{2}(45^{ \circ }+\alpha)+1}=sin2\alpha\)

Za rozwiązania dziękuję

[irena]

1.
a)
\(sin^4\alpha-cos^4\alpha=(sin^2\alpha+cos^2\alpha)(sin^2\alpha-cos^2\alpha)=sin^2\alpha-cos^2\alpha=\\=sin^2\alpha-(1-sin^2\alpha)=sin^2\alpha-1+sin^2\alpha=2sin^2\alpha-1\)

[jola]

\(L=\sin^4 \alpha -\cos^4 \alpha =(\sin^2 \alpha-\cos^2 \alpha )(\sin^2 \alpha +\cos^ \alpha )=\sin^2 \alpha -1+\sin^2 \alpha=2\sin^2 \alpha -1=P\)

[irena]

b)
\(\frac{sin5\alpha+sin3\alpha}{cos5\alpha+cos3\alpha}=\frac{2sin4\alpha cos\alpha}{2cos4\alpha cos\alpha}=\frac{sin4\alpha}{cos4\alpha}=tg4\alpha\)

[irena]

c)
\(tg\alpha\cdot sin2\alpha=\frac{sin\alpha}{cos\alpha}\cdot2sin\alpha cos\alpha=2sin^2\alpha\)

[irena]

d)
\(tg(45^0+\alpha)\cdot tg(45^0-\alpha)=tg(45^0+\alpha)\cdot ctg(90^0-45^0+\alpha)=tg(45^0+\alpha)\cdot ctg(45^0+\alpha)=1\)

[irena]

2.
a)
\(\frac{cos\alpha-cos\beta}{sin\alpha+sin\beta}=\frac{-2sin{\frac{\alpha+\beta}{2}}sin{\frac{\alpha-\beta}{2}}}{2sin{\frac{\alpha+\beta}{2}}cos{\frac{\alpha-\beta}{2}}}=-tg{\frac{\alpha-\beta}{2}}=tg{\frac{\beta-\alpha}{2}}\)

[irena]

b)
\(\frac{cos(\alpha-\beta)+cos(\alpha-\beta)}{sin(\alpha-\beta)+sin(\alpha+\beta)}=\frac{2cos\alpha cos(-\beta)}{2sin\alpha cos(-\beta)}=ctg\alpha\)

[jola]

zad 2c
\(45^ \circ + \alpha \neq 90^ \circ +k \cdot 180^ \circ \ \ \Rightarrow\ \ \alpha \neq 45^ \circ +k \cdot 180^ \circ \\ L= \frac{ \frac{\sin^2(45^ \circ \alpha )-\cos^2 (45^ \circ +\alpha0 }{\cos^2(45^ \circ + \alpha) } }{ \frac{\sin^2(45^ \circ + \alpha )+\cos^2 (45^ \circ \alpha )}{\cos^2(45^ \circ + \alpha) } } =-\cos(90^ \circ +2 \alpha )=\sin 2 \alpha =P\)