Question

Urgent ! Ajutati-ma !


$\sin \: a \: = \frac{2}{3}$

$\cos b= - \frac{1}{3}$
a , b ∈ ( pi / 2 , 2 )

Calculati :
sin ( a+b)
sin (a-b)
cos (a+b)
cos (a-b )

P.S : sin ( a+b) este deja calculat
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Answer 1

$\it b\in \Big(\dfrac{\pi}{2},\ \pi\Big) \Rightarrow sinb=\sqrt{1-cos^2b} =\sqrt{1-\dfrac{1}{9}}=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt2}{3}\\ \\ \\ a\in \Big(\dfrac{\pi}{2},\ \pi\Big) \Rightarrow cosa=-\sqrt{1-sin^2a} =-\sqrt{1-\dfrac{4}{9}}=-\sqrt{\dfrac{5}{9}}=-\dfrac{\sqrt5}{3}$

$\it sin(a+b)=sinacosb+sinbcosa=\dfrac{2}{3}\cdot(-\dfrac{1}{3})+\dfrac{2\sqrt2}{3}\cdot(-\dfrac{\sqrt5}{3})=-\dfrac{2}{9}(1+\sqrt{10})$