Question

помогите кто сможет решить задачу 3

Answer 1

$Sin\alpha=\dfrac{1}{3}\\\\0<\alpha<\frac{\pi }{2} \ \Rightarrow \ Cos\alpha >0\\\\Cos\alpha=\sqrt{1-Sin^{2}\alpha}=\sqrt{1-\Big(\dfrac{1}{3}\Big)^{2}}=\sqrt{1-\dfrac{1}{9} } =\sqrt{\dfrac{8}{9} }=\dfrac{2\sqrt{2} }{3}\\\\\\Sin\Big(\dfrac{\pi }{6}+\alpha\Big)=Sin\dfrac{\pi }{6}\cdot Cos\alpha+Cos\dfrac{\pi }{6}\cdot Sin\alpha =\dfrac{1}{2}\cdot \dfrac{2\sqrt{2} }{3}+\dfrac{\sqrt{3}}{2}\cdot \dfrac{1}{3}=$

$=\dfrac{\sqrt{2} }{3}+\dfrac{\sqrt{3} }{6} =\boxed{\dfrac{2\sqrt{2}+\sqrt{3}}{6}}$