[tex]sin\alpha = \frac{9}{14}\\cos\alpha = ?[/tex]
Wiemy, że:
[tex]sin^{2}\alpha + cos^{2}\alpha = 1\\\\cos^{2}\alpha = 1 - sin^{2}\alpha = 1 - (\frac{9}{14})^{2} = \frac{196}{196}-\frac{81}{196} = \frac{115}{196}\\\\cos\alpha = \sqrt_\frac{115}{196}} = \frac{\sqrt{115}}{\sqrt{196}} = \frac{\sqrt{115}}{14}[/tex]
