Question

the real number t corresponds to the point p on the unit circle. identify the coordinates of point p. then evaluate the six trigonometric functions of t. write your answer as a simplified fraction, if necessary. rationalize the denominator, if necessary.

$t = \frac{\pi}{4}$

part 1 of 7

$p(x,y)=$

Explanation:

Step1: Recall unit - circle coordinates formula

For a real number \(t\) on the unit circle \(x = \cos t\) and \(y=\sin t\). Given \(t = \frac{\pi}{4}\), we know that \(\cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\).
So the coordinates of point \(P(x,y)\) are \((\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\).

Step2: Evaluate sine function

\(\sin t=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\)

Step3: Evaluate cosine function

\(\cos t=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\)

Step4: Evaluate tangent function

\(\tan t=\frac{\sin t}{\cos t}\), substituting \(t = \frac{\pi}{4}\), we get \(\tan\frac{\pi}{4}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1\)

Step5: Evaluate cosecant function

\(\csc t=\frac{1}{\sin t}\), substituting \(t=\frac{\pi}{4}\), we have \(\csc\frac{\pi}{4}=\frac{1}{\frac{\sqrt{2}}{2}}=\sqrt{2}\)

Step6: Evaluate secant function

\(\sec t=\frac{1}{\cos t}\), substituting \(t = \frac{\pi}{4}\), we get \(\sec\frac{\pi}{4}=\frac{1}{\frac{\sqrt{2}}{2}}=\sqrt{2}\)

Step7: Evaluate cotangent function

\(\cot t=\frac{\cos t}{\sin t}\), substituting \(t=\frac{\pi}{4}\), we have \(\cot\frac{\pi}{4}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1\)

Answer:

\(P(x,y)=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\); \(\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\); \(\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\); \(\tan\frac{\pi}{4}=1\); \(\csc\frac{\pi}{4}=\sqrt{2}\); \(\sec\frac{\pi}{4}=\sqrt{2}\); \(\cot\frac{\pi}{4}=1\)