Find the exact value of each of the six trigonometric functions of \( \theta \), if \( (1,1) \) is a point on the terminal side of angle \( \theta \). \( \sin \theta=\square \) \( ( \) Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression. Rationalize all denominators.)

To find the exact values of the six trigonometric functions of the angle \( \theta \) given that the point \( (1, 1) \) is on the terminal side of \( \theta \), we start by determining the radius \( r \) of the corresponding right triangle formed by the point and the origin. The coordinates of the point are \( (x, y) = (1, 1) \). The radius \( r \) is calculated using the Pythagorean theorem: \[ r = \sqrt{x^2 + y^2} = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}. \] Now we can find the six trigonometric functions: 1. **Sine**: \[ \sin \theta = \frac{y}{r} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \quad \text{(after rationalizing the denominator)}. \] 2. **Cosine**: \[ \cos \theta = \frac{x}{r} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \quad \text{(after rationalizing the denominator)}. \] 3. **Tangent**: \[ \tan \theta = \frac{y}{x} = \frac{1}{1} = 1. \] 4. **Cosecant** (the reciprocal of sine): \[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \quad \text{(after rationalizing the denominator)}. \] 5. **Secant** (the reciprocal of cosine): \[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \quad \text{(after rationalizing the denominator)}. \] 6. **Cotangent** (the reciprocal of tangent): \[ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{1} = 1. \] Now we can summarize the values of the six trigonometric functions: \[ \sin \theta = \frac{\sqrt{2}}{2}, \] \[ \cos \theta = \frac{\sqrt{2}}{2}, \] \[ \tan \theta = 1, \] \[ \csc \theta = \sqrt{2}, \] \[ \sec \theta = \sqrt{2}, \] \[ \cot \theta = 1. \]