Question

sa se calculeze cos(a+b) daca tg a =3/4,cos b=5/13,a,b€(0,π/2)

Answer 1

[tex]\sin^2 b+\cos ^2 b=1\\ \sin^2 b=1-\cos^2 b\\ \sin^2 b=1-\dfrac{25}{169}\\ \\ \sin^2 b=\dfrac{144}{169}\\ \\ \sin b=\pm \dfrac{12}{13}\\ b\in \left(0,\dfrac{\pi }{2} \right)\Rightarrow \sin b=\dfrac{12}{13}\\ tg\ a=\dfrac{3}{4}\\ \dfrac{\sin a}{\cos a}=\dfrac{3}{4}\Rightarrow \sin a=\dfrac{3}{4}\cdot \cos a\\ \sin^2 a+\cos^2 a=1\\ \dfrac{9}{16}\cos^2a +\cos^2 a=1\\ \\ \dfrac{25}{16}\cos^2 a=1\\ \cos^2 a=\dfrac{16}{25}\Rightarrow \cos a=\pm \dfrac{4}{5} [/tex]
[tex]a\in \left(0,\dfrac{\pi}{2} \right) \Rightarrow \cos a=\dfrac{4}{5}\\ \sin a=\dfrac{3}{4}\cdot \sin a=\dfrac{3}{5}\\ \\ cos(a+b)=\cos a\cdot \cos b-\sin a\cdot \sin b=\dfrac{4}{5}\cdot \dfrac{5}{13}-\dfrac{3}{5}\cdot \dfrac{12}{13}=\\ =\dfrac{4}{13}-\dfrac{36}{65}=\dfrac{20-36}{65}= \boxed{-\dfrac{16}{65}} [/tex]

Answer 2

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