Given a point on the unit circle, determine the coordinates and verify the trigonometric identities.

Let’s consider a point $P(\cos\theta, \sin\theta)$ on the unit circle where $\theta = \frac{5\pi}{6}$. To find the coordinates and verify trigonometric identities:

First, we calculate the coordinates:

$P = (\cos \frac{5\pi}{6}, \sin \frac{5\pi}{6})$

Using the unit circle, we know:

$\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$

$\sin \frac{5\pi}{6} = \frac{1}{2}$

Thus, the coordinates are:

$P = \left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$

Next, we verify the Pythagorean identity:

$\cos^2 \theta + \sin^2 \theta = 1$

Substituting in the values, we get:

$\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1$

Which confirms that the point lies on the unit circle.

Let’s consider a point $P(cos heta, sin heta)$ on the unit circle where $ heta = frac{3pi}{4}$. To find the coordinates and verify trigonometric identities:

First, we calculate the coordinates:

$P = (cos frac{3pi}{4}, sin frac{3pi}{4})$

Using the unit circle, we know:

$cos frac{3pi}{4} = -frac{sqrt{2}}{2}$

$sin frac{3pi}{4} = frac{sqrt{2}}{2}$

Thus, the coordinates are:

$P = left(-frac{sqrt{2}}{2}, frac{sqrt{2}}{2}
ight)$

Next, we verify the Pythagorean identity:

$cos^2 heta + sin^2 heta = 1$

Substituting in the values, we get:

$left(-frac{sqrt{2}}{2}
ight)^2 + left(frac{sqrt{2}}{2}
ight)^2 = frac{2}{4} + frac{2}{4} = 1$

Which confirms that the point lies on the unit circle.

Consider a point $P(cos heta, sin heta)$ on the unit circle where $ heta = frac{7pi}{6}$. Find the coordinates and verify the Pythagorean identity.

First, we calculate the coordinates:

$cos frac{7pi}{6} = -frac{sqrt{3}}{2}$

$sin frac{7pi}{6} = -frac{1}{2}$

So, coordinates are:

$P = left(-frac{sqrt{3}}{2}, -frac{1}{2}
ight)$

Verify the identity:

$left(-frac{sqrt{3}}{2}
ight)^2 + left(-frac{1}{2}
ight)^2 = 1$

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