if \\( \\tan \\theta = \\frac { 2 } { 3 }, 0 < \\theta < \\frac { \\pi } { 2 } \\), find \\( \\sin \\frac {…

Explanation:

Step1: Find (\cos\theta)

Given (\tan\theta=\frac{2}{3}=\frac{\text{opposite}}{\text{adjacent}}). By the Pythagorean theorem, the hypotenuse (r = \sqrt{2^{2}+3^{2}}=\sqrt{4 + 9}=\sqrt{13}). Then (\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{3}{\sqrt{13}}=\frac{3\sqrt{13}}{13}).

Step2: Use the half - angle formula for sine

The half - angle formula for sine is (\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}). Since (0\lt\theta\lt\frac{\pi}{2}), then (0\lt\frac{\theta}{2}\lt\frac{\pi}{4}), and (\sin\frac{\theta}{2}\gt0). Substitute (\cos\theta=\frac{3\sqrt{13}}{13}) into the formula: [ \begin{align*} \sin\frac{\theta}{2}&=\sqrt{\frac{1-\frac{3\sqrt{13}}{13}}{2}}\ &=\sqrt{\frac{\frac{13 - 3\sqrt{13}}{13}}{2}}\ &=\sqrt{\frac{13- 3\sqrt{13}}{26}} \end{align*} ]

Answer:

(\sqrt{\frac{13 - 3\sqrt{13}}{26}})