Homework

Find the exact values of sin, cos, and tan for 7π/6 using the unit circle

To find the exact values of $ \sin, \cos, $ and $ \tan $ for $ \frac{7\pi}{6} $ using the unit circle, we need to determine the coordinates of the point corresponding to this angle.

Since $ \frac{7\pi}{6} $ is in the third quadrant, both sine and cosine values will be negative:

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

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

Now, using the identity $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $:

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

Find the tan values of the unit circle at specific angles

To find the $\tan$ values of the unit circle at specific angles, we can use the fact that $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$:

1. At $\theta = \frac{\pi}{4}$, $$ \tan(\frac{\pi}{4}) = \frac{\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{4})} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$

2. At $\theta = \frac{3\pi}{4}$, $$ \tan(\frac{3\pi}{4}) = \frac{\sin(\frac{3\pi}{4})}{\cos(\frac{3\pi}{4})} = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1 $$

3. At $\theta = \pi$, $$ \tan(\pi) = \frac{\sin(\pi)}{\cos(\pi)} = \frac{0}{-1} = 0 $$

Find the coordinates of the point on the unit circle corresponding to the angle 7π/6

To find the coordinates on the unit circle for the angle $\frac{7\pi}{6}$, we use the unit circle properties:

The unit circle coordinates $(x, y)$ for an angle $\theta$ are $(\cos(\theta), \sin(\theta))$.

For $\theta = \frac{7\pi}{6}$:

$$ x = \cos\left(\frac{7\pi}{6}\right) $$

$$ y = \sin\left(\frac{7\pi}{6}\right) $$

Using trigonometric identities:

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

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

Therefore, the coordinates are:

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

Find the values of sin(θ), cos(θ), and tan(θ) for θ = π/4 using the unit circle

To find the values of $\sin(\theta)$, $\cos(\theta)$, and $\tan(\theta)$ for $\theta = \frac{\pi}{4}$ using the unit circle, we use the following:

On the unit circle, at $\theta = \frac{\pi}{4}$, the coordinates are: $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.

So,

$$ \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} $$

$$ \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} $$

$$ \tan(\frac{\pi}{4}) = \frac{\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{4})} = 1 $$

Find the values of sin(θ) at specific angles on the unit circle

To find the values of $ \sin(\theta) $ at specific angles on the unit circle, we can use the known values for common angles:

At $ \theta = 0 $, $$ \sin(0) = 0 $$

At $ \theta = \frac{\pi}{2} $, $$ \sin\left(\frac{\pi}{2}\right) = 1 $$

At $ \theta = \pi $, $$ \sin(\pi) = 0 $$

At $ \theta = \frac{3\pi}{2} $, $$ \sin\left(\frac{3\pi}{2}\right) = -1 $$

At $ \theta = 2\pi $, $$ \sin(2\pi) = 0 $$

Prove that tan(theta) sec(theta) = sin(theta) where theta is an angle in the unit circle

We start with the definitions of the trigonometric functions on the unit circle.

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

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$$ \sec(\theta) = \frac{1}{\cos(\theta)} $$

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Multiplying these two expressions, we have:

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$$ \tan(\theta) \sec(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \cdot \frac{1}{\cos(\theta)} $$

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Simplifying, we get:

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

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Since $ \cos^2(\theta) \cos(\theta) = \cos(\theta) $, we have:

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$$ \tan(\theta) \sec(\theta) = \sin(\theta) \cdot \frac{1}{\cos^2(\theta)} = \sin(\theta) $$

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Thus, it is proven that:

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$$ \boxed{\tan(\theta) \sec(\theta) = \sin(\theta)} $$

Find the exact values of the coordinates of the point where the unit circle intersects the positive x-axis

The unit circle is defined by the equation:

$$ x^2 + y^2 = 1 $$

The positive x-axis means $ y = 0 $. Substituting $ y = 0 $ into the equation gives:

$$ x^2 + 0^2 = 1 $$

Simplifying, we find:

$$ x^2 = 1 $$

Taking the positive square root (since we are on the positive x-axis), we get:

$$ x = 1 $$

Thus, the coordinates of the point are:

$$ (1, 0) $$

Identify the sine value of an angle corresponding to $3\pi/4$

We start by noting that $ \frac{3\pi}{4} $ is in the second quadrant of the unit circle.

In the second quadrant, the sine value is positive, so we have:

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

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

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

Find the exact values of arcsec(2)

To find the exact value of $ \text{arcsec}(2) $, we need to determine the angle $ \theta $ such that $ \sec(\theta) = 2 $ and $ \theta $ lies within the range of secant

Find $ sin(θ) $ and $ cos(θ) $ for θ on the unit circle

To find $ \sin(\theta) $ and $ \cos(\theta) $ when $ \theta $ is on the unit circle:

Recall the unit circle definition: the unit circle is a circle with a radius of 1 centered at the origin. Therefore, if $ (x, y) $ is a point on the unit circle corresponding to the angle $ \theta $, then:

$$ \cos(\theta) = x $$

$$ \sin(\theta) = y $$

For example, at $ \theta = \frac{\pi}{4} $, we have:

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

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