Diketahui sin a = 1/5 √13, a sudut lancip. nilai cos 2a adalah.... Question from @silvires09

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Diketahui sin a = 1/5 √13, a sudut lancip. nilai cos 2a adalah

renikartikareka Sin a = 1/5 √13
sin² a = (1/5 √13)²
= 1/25 . 13 = 13/25

cos 2a = 1 - 2 sin²a
= 1 - 2 . 13/25
= 25/25 - 26/25
= -1/25

13 votes Thanks 34

izharulhaq78 $\sin a=\frac{\sqrt{13}}{5}\\\\a=\sqrt{13}\\c=5\\b=\sqrt{c^2-a^2}=\sqrt{5^2-\sqrt{13}^2}=\sqrt{25-13}=\sqrt{12}\\\\\cos2a=\cos^2a-\sin^2a\\\cos2a=(\cos a)^2-(\sin a)^2\\\cos2a=(\frac{b}{c})^2-(\frac{a}{c})^2\\\cos2a=(\frac{\sqrt{12}}{5})^2-(\frac{\sqrt{13}}{5})^2\\\cos2a=(\frac{12}{25})-(\frac{13}{25})\\\cos2a=\frac{12}{25}-\frac{13}{25}\\\cos2a=-\frac{1}{25}$

6 votes Thanks 13

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