Question

Ajutor va rog frumos

Answer 1

4.

$sin²a+cos²a=1=>sin²a+ \frac{25 }{169}=1 = > sin²a= \frac{169-25}{169} => sin²a= \frac{144}{169} =>sin a= \frac{12}{13}$

$sin2a = 2 \times sin \: a \: \times cos \: a = 2 \times \frac{5}{13} \times \frac{12}{13} = \frac{120}{13}$

$cos2a = cos^{2} a - sin ^{2} a = \frac{25}{169} - \frac{144}{169} = - \frac{119}{169}$

Answer 2

Explicație pas cu pas:

${ \sin}^{2}a + { \cos}^{2}a = 1 \iff { \sin}^{2}a = 1 - { \cos}^{2}a$

${ \sin}^{2}a = 1 - {\Big( - \frac{5}{13}\Big)}^{2} = 1 - \frac{25}{169} = \frac{144}{169}$

$a \in \Big( \frac{\pi}{2} ;\pi\Big) \implies \: cadranul \: 2$

$\implies \bf \sin a = \frac{12}{13}$

$\sin(2a) = 2 \sin a \cdot \cos a = 2 \cdot \frac{12}{13} \cdot \Big( - \frac{5}{13}\Big) \\ \implies \bf \sin(2a) = - \frac{120}{169}$

$\cos(2a) = 1 - 2\sin^{2} (a) = 1 - 2 \times \frac{144}{169} = \frac{169 - 288}{169} \\ \implies \bf \cos(2a) = - \frac{119}{169}$