Odpowiedź:
[tex]0 = \frac{1}{\sin\alpha}-\frac{1}{\cos\alpha} = \frac{\cos\alpha-\sin\alpha}{\sin\alpha\cos\alpha} \rightarrow \cos\alpha-\sin\alpha=0 \rightarrow \cos\alpha=\sin\alpha \rightarrow \tan\alpha=1[/tex]
α jest kątem ostrym, jedyny kąt ostry spełniający to równanie to
α = 45°
[tex]\huge\boxed{\alpha = 45^{o}}[/tex]
[tex]\frac{1}{sin\alpha} - \frac{1}{cos\alpha}=0\\\\\frac{1}{sin\alpha} = \frac{1}{cos\alpha}\\\\sin\alpha = cos\alpha \ \ \ |()^{2}\\\\sin^{2}\alpha = cos^{2} \ \alpha\\\\Z \ jedynki \ trygonometrycznej:\\\\si^{2}\alpha + cos^{2}\alpha = 1 \ \ \rightarrow \ \ cos^{2}\alpha = 1-sin^{2}\alpha\\\\sin^{2}\alpha =1-sin^{2}\alpha\\\\sin^{2}\alpha + sin^{2}\alpha = 1\\\\2sin^{2}\alpha = 1 \ \ \ /:2\\\\sin^{2}\alpha = \frac{1}{2}\\\\0^{o} < \alpha < 90^{o}[/tex]
[tex]sin\alpha = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2} \ \ \rightarrow \ \boxed{ \alpha = 45^{o}}[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Odpowiedź:
[tex]0 = \frac{1}{\sin\alpha}-\frac{1}{\cos\alpha} = \frac{\cos\alpha-\sin\alpha}{\sin\alpha\cos\alpha} \rightarrow \cos\alpha-\sin\alpha=0 \rightarrow \cos\alpha=\sin\alpha \rightarrow \tan\alpha=1[/tex]
α jest kątem ostrym, jedyny kąt ostry spełniający to równanie to
α = 45°
Verified answer
[tex]\huge\boxed{\alpha = 45^{o}}[/tex]
[tex]\frac{1}{sin\alpha} - \frac{1}{cos\alpha}=0\\\\\frac{1}{sin\alpha} = \frac{1}{cos\alpha}\\\\sin\alpha = cos\alpha \ \ \ |()^{2}\\\\sin^{2}\alpha = cos^{2} \ \alpha\\\\Z \ jedynki \ trygonometrycznej:\\\\si^{2}\alpha + cos^{2}\alpha = 1 \ \ \rightarrow \ \ cos^{2}\alpha = 1-sin^{2}\alpha\\\\sin^{2}\alpha =1-sin^{2}\alpha\\\\sin^{2}\alpha + sin^{2}\alpha = 1\\\\2sin^{2}\alpha = 1 \ \ \ /:2\\\\sin^{2}\alpha = \frac{1}{2}\\\\0^{o} < \alpha < 90^{o}[/tex]
[tex]sin\alpha = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2} \ \ \rightarrow \ \boxed{ \alpha = 45^{o}}[/tex]