Unit Circle

Find the coordinates of the point on the unit circle where the angle is 3π/4 radians

To find the coordinates of the point on the unit circle where the angle is $\frac{3\pi}{4}$ radians, we use the unit circle definition:

The coordinates are given by:

$$ (\cos\theta, \sin\theta) $$

For $ \theta = \frac{3\pi}{4} $:

$$ \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} $$

$$ \sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} $$

Therefore, the coordinates are:

$$ \left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) $$

Determine the exact values of the trigonometric functions for angle 7π/6

To determine the exact values of the trigonometric functions for angle $ \frac{7\pi}{6} $, we follow these steps:

1. Recognize that $ \frac{7\pi}{6} $ is in the third quadrant.
2. Calculate the reference angle:
$$ \pi – \frac{7\pi}{6} = \frac{\pi}{6} $$.

The sine and cosine values in the third quadrant are negative:

$$ \sin \left( \frac{7\pi}{6} \right) = -\sin \left( \frac{\pi}{6} \right) = -\frac{1}{2} $$

$$ \cos \left( \frac{7\pi}{6} \right) = -\cos \left( \frac{\pi}{6} \right) = -\frac{\sqrt{3}}{2} $$

3. Calculate the tangent value:

$$ \tan \left( \frac{7\pi}{6} \right) = \frac{\sin \left( \frac{7\pi}{6} \right)}{\cos \left( \frac{7\pi}{6} \right)} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} $$

Find the coordinates of a point on the unit circle where the angle is given by theta = (3/4) * pi

The unit circle

Identify the quadrant of the unit circle

To identify the quadrant where the angle $ \theta = \frac{3\pi}{4} $ lies, we need to examine the unit circle.

$$ \frac{3\pi}{4} $$ is in radians.

The angle $ \theta = \frac{3\pi}{4} $ is less than $ \pi $ but more than $ \frac{\pi}{2} $.

Therefore, $ \theta = \frac{3\pi}{4} $ is in the second quadrant.

Determine the quadrant of various points on the unit circle

To determine the quadrant of a point on the unit circle, consider the signs of the x and y coordinates:

Quadrant I: Both coordinates are positive ($x > 0$, $y > 0$)

Quadrant II: x is negative, y is positive ($x < 0$, $y > 0$)

Quadrant III: Both coordinates are negative ($x < 0$, $y < 0$)

Quadrant IV: x is positive, y is negative ($x > 0$, $y < 0$)

Calculate cos(π/4) using the unit circle

To calculate $ \cos(\frac{\pi}{4}) $ using the unit circle, we look at the angle $ \frac{\pi}{4} $ on the unit circle.

At this angle, both sine and cosine values are equal.

Thus, the value of $ \cos(\frac{\pi}{4}) $ is:

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

How to find tan on unit circle?

To find the tangent of an angle $ \theta $ on the unit circle, you need to know the coordinates of the point where the terminal side of the angle intersects the unit circle. The coordinates are given by $ ( \cos(\theta), \sin(\theta) ) $.

The tangent of the angle $ \theta $ is given by:

$$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$

For example, if $ \theta = \frac{\pi}{4} $, then:

$$ \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

and

$$ \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} $$

so

$$ \tan \left( \frac{\pi}{4} \right) = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

Prove that tan(2θ) = 2tan(θ) / (1 – tan^2(θ)) using the unit circle

Consider a point on the unit circle at an angle $ \theta $. Using the double-angle identities, we can write:

$$ \sin(2\theta) = 2\sin(\theta)\cos(\theta) $$

and

$$ \cos(2\theta) = \cos^2(\theta) – \sin^2(\theta) $$

Thus,

$$ \tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta) – \sin^2(\theta)} $$

Using the fact that $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $, let

$$ t = \tan(\theta) $$

Then

$$ \sin(\theta) = \frac{t}{\sqrt{1 + t^2}} $$

and

$$ \cos(\theta) = \frac{1}{\sqrt{1 + t^2}} $$

Substituting these into the double-angle formula gives:

$$ \tan(2\theta) = \frac{2 \cdot \frac{t}{\sqrt{1 + t^2}} \cdot \frac{1}{\sqrt{1 + t^2}}}{\left(\frac{1}{\sqrt{1 + t^2}}\right)^2 – \left(\frac{t}{\sqrt{1 + t^2}}\right)^2} $$

Which simplifies to:

$$ \tan(2\theta) = \frac{2t}{1 – t^2} $$

Determine the cosine of the angle t on the unit circle when the sine of t is 1/2

To find the cosine of the angle $t$ on the unit circle when the sine of $t$ is $\frac{1}{2}$, we can use the Pythagorean identity:

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$$ \sin^2(t) + \cos^2(t) = 1 $$

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Given that $\sin(t) = \frac{1}{2}$, we substitute and solve for $\cos(t)$:

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$$ \left(\frac{1}{2}\right)^2 + \cos^2(t) = 1 $$

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$$ \frac{1}{4} + \cos^2(t) = 1 $$

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$$ \cos^2(t) = 1 – \frac{1}{4} $$

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$$ \cos^2(t) = \frac{3}{4} $$

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Therefore, $\cos(t)$ can be:

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$$ \cos(t) = \pm\frac{\sqrt{3}}{2} $$

Find the angle θ in the unit circle where the sum of sin(θ) and cos(θ) equals 15

To find the angle $ \theta $ where the sum of $ \sin(\theta) $ and $ \cos(\theta) $ equals 1.5, we start with the equation:

$$ \sin(\theta) + \cos(\theta) = 1.5 $$

We can use the Pythagorean identity:

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

Let’s square both sides of the original equation:

$$ (\sin(\theta) + \cos(\theta))^2 = 1.5^2 $$

This gives us:

$$ \sin^2(\theta) + 2\sin(\theta)\cos(\theta) + \cos^2(\theta) = 2.25 $$

Using the Pythagorean identity:

$$ 1 + 2\sin(\theta)\cos(\theta) = 2.25 $$

Therefore:

$$ 2\sin(\theta)\cos(\theta) = 1.25 $$

Which simplifies to:

$$ \sin(2\theta) = 1.25 $$

However, we know that the range of $ \sin(2\theta) $ is between -1 and 1, so no such $ \theta $ exists.