Odpowiedź:
[tex]\huge\boxed {\huge\boxed {~~2cos^{2}\alpha -1=-\frac{3}{5} ~~}}[/tex]
Szczegółowe wyjaśnienie:
Korzystamy ze wzorów:
dane z treści zadania:
szukane:
[tex]tg\alpha =\dfrac{sin\alpha }{cos\alpha } ~~\land~~tg\alpha =2\\\\~~~~~~~~~~~~~~~~~~\Downarrow\\\\\dfrac{sin\alpha }{cos\alpha } =2~~\mid \cdot cos\alpha \\\\\boxed {~~sin\alpha =2cos\alpha ~~}\\\\sin^{2}\alpha +cos^{2}\alpha =1~~\land~~sin\alpha =2cos\alpha \\\\~~~~~~~~~~~~~~~~~~~~~~~~\Downarrow\\\\(2cos\alpha )^{2}+cos^{2}\alpha =1\\\\4cos^{2}\alpha +cos^{2}\alpha =1\\\\5cos^{2}\alpha =1~~\mid \div 5\\\\\boxed {~~cos^{2}\alpha =\dfrac{1}{5} ~~}[/tex]
[tex]2cos^{2}\alpha -1=?~~\land~~cos^{2}\alpha =\dfrac{1}{5}\\ \\~~~~~~~~~~~~~~~~~~~~~~\Downarrow \\\\2\cdot \dfrac{1}{5} -1=\dfrac{2}{5} -1=\dfrac{2}{5} -\dfrac{5}{5} =-\dfrac{3}{5}[/tex]
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Verified answer
Odpowiedź:
[tex]\huge\boxed {\huge\boxed {~~2cos^{2}\alpha -1=-\frac{3}{5} ~~}}[/tex]
Szczegółowe wyjaśnienie:
Korzystamy ze wzorów:
Rozwiązanie:
dane z treści zadania:
szukane:
[tex]tg\alpha =\dfrac{sin\alpha }{cos\alpha } ~~\land~~tg\alpha =2\\\\~~~~~~~~~~~~~~~~~~\Downarrow\\\\\dfrac{sin\alpha }{cos\alpha } =2~~\mid \cdot cos\alpha \\\\\boxed {~~sin\alpha =2cos\alpha ~~}\\\\sin^{2}\alpha +cos^{2}\alpha =1~~\land~~sin\alpha =2cos\alpha \\\\~~~~~~~~~~~~~~~~~~~~~~~~\Downarrow\\\\(2cos\alpha )^{2}+cos^{2}\alpha =1\\\\4cos^{2}\alpha +cos^{2}\alpha =1\\\\5cos^{2}\alpha =1~~\mid \div 5\\\\\boxed {~~cos^{2}\alpha =\dfrac{1}{5} ~~}[/tex]
[tex]2cos^{2}\alpha -1=?~~\land~~cos^{2}\alpha =\dfrac{1}{5}\\ \\~~~~~~~~~~~~~~~~~~~~~~\Downarrow \\\\2\cdot \dfrac{1}{5} -1=\dfrac{2}{5} -1=\dfrac{2}{5} -\dfrac{5}{5} =-\dfrac{3}{5}[/tex]