sin t=cos t=tan t= csc t=sec t = cot t = ### Trigonometric Functions on the Unit CircleThe real number \( t \) corresponds to the point \( P \left( -\frac{5}{6}, \frac{\sqrt{11}}{6} \right) \) on the unit circle. Evaluate the six trigonometric functions of \( t \). Write your answer as a simplified fraction, if necessary.#### Graph DescriptionThe unit circle is illustrated with its center at the origin \((0,0)\) and radius equal to 1. The circle is defined by the equation \( x^2 + y^2 = 1 \). The point \( P \left( -\frac{5}{6}, \frac{\sqrt{11}}{6} \right) \) is marked on the circle, which lies in the second quadrant where the x-coordinate is negative and the y-coordinate is positive.#### Required CalculationPerform the following calculations:**Part 1 of 6:**Calculate \(\sin t \)\[ \sin t = \sqrt{\frac{11}{6}} \ (\frac{\sqrt{11}}{6})\]Please proceed with the next calculations for the complete evaluation of the trigonometric functions.### Note:To submit your answers and verify if they are correct, please use the given user interface. Continue for more parts...

The given point on the circle is (-5/6, -√11/6)

The real number / corresponds to the point P fraction, if necessary. O: 2 0/6 X 2 x + y = 1 Part 1 of 6 sin t = Continue 6,385 S JUN 6 11 on the unit circle. Evaluate the six trigonometric functions of t. Write your answer as a simplified 6 Sub Ⓒ2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Ces NOO O tv MacBook Air