Sprawdz tożsamosc

[marco_te]

1) \(\displaystyle{ \sin ^{2} \alpha-\sin^{2}\beta= \cos^{2}\beta-\cos^{2}\alpha}\)
2) \(\displaystyle{ \frac{\sin^{3}\alpha}{\cos\alpha-\cos^{3}\alpha}=tg\alpha}\)
3) \(\displaystyle{ \frac{1-\sin\alpha}{\sin\alpha}*\frac{1+sin^{2}}{\cos\alpha}=\ctg\alpha}\)
4) \(\displaystyle{ \frac{\sin\alpha}{1-\cos^{2}}-\frac{1+\cos^{2}}{\sin\alpha}=0}\)
5) \(\displaystyle{ (\sin\alpha+\tg\alpha)*(\cos\alpha+\ctg\alpha)=(1+\sin\alpha)*(1+\cos\alpha)}\)
6) \(\displaystyle{ \tg\alpha+\frac{\cos\alpha}{1+\sin\alpha}=\frac{1}{\cos\alpha}}\)

[Mat.Monia]

1)\(\displaystyle{ sin ^{2} \alpha +cos ^{2}\beta=1}\), więc \(\displaystyle{ sin ^{2} \alpha =1-cos ^{2} \beta}\)
stąd
\(\displaystyle{ L= 1-cos ^{2}\alpha-(1-cos ^{2}\beta)=cos ^{2}\beta-cos ^{2} \alpha=P}\)

2)\(\displaystyle{ \frac{sin ^{3} \alpha }{cos \alpha -cos ^{3} \alpha }= \frac{sin ^{3} }{cos \alpha (1-cos^{2})}= \frac{sin ^{3} \alpha }{cos \alpha *sin ^{2} \alpha } =sin \alpha /cos \alpha =tg \alpha =p}\)

[marco_te]

a reszte moglby ktos pomuc??;>

[agulka1987]

6.

\(\displaystyle{ tg\alpha + \frac{cos\alpha}{1+sin\alpha} = \frac{sin\alpha}{cos\alpha} + \frac{cos\alpha}{1+sin\alpha} = \frac{sin\alpha+sin^2\alpha+cos^2\alpha }{cos\alpha(1+sin\alpha)} = \frac{sin\alpha + 1 - cos^2\alpha+cos^2\alpha}{cos\alpha(1+sin\alpha)} = \frac{sin\alpha+1}{cos\alpha(1+sin\alpha)} = \frac{1}{cos\alpha}}\)-- 17 kwietnia 2009, 14:48 --5.

\(\displaystyle{ (sin\alpha + tg\alpha)(cos\alpha+ctg\alpha) = (sin\alpha+ \frac{sin\alpha}{cos\alpha})(cos\alpha+ \frac{cos\alpha}{sin\alpha}) = \frac{sin\alpha(1+cos\alpha)}{cos\alpha} \cdot \frac{cos\alpha(1+sin\alpha)}{sin\alpha} = (1+cos\alpha)(1+sin\alpha)}\)

[matshadow]

4) jest źle.
\(\displaystyle{ 1-\cos^2\alpha=\sin^2\alpha}\)
\(\displaystyle{ \frac{1}{\sin\alpha}=\frac{1+\cos^2\alpha}{\sin\alpha}}\)
Co nie jest tożsamością