Problem 15 - Algebra 2 and Trigonometry

Question

In \(13-20, P\) is a point on the terminal side of an angle in standard position with measure \(\theta\) and on a circle with center at the origin and radius \(r .\) For each point \(P,\) find: a. \(r\) b. \(\cos \theta\) c. \(\sin \theta\) d. \(\tan \theta\) $$ (-1,-1) $$

Step-by-Step Solution

Answer

a. \( r = \sqrt{2} \), b. \( \cos\theta = -\frac{\sqrt{2}}{2} \), c. \( \sin\theta = -\frac{\sqrt{2}}{2} \), d. \( \tan\theta = 1 \).

Step 1: Determine the Radius

The point \( P(-1,-1) \) lies on a circle centered at the origin. We use the distance formula to find the radius, \( r \), which is also the same as the hypotenuse of the right triangle formed. Use \( r = \sqrt{x^2 + y^2} \). Calculate:\[ r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \].Thus, the radius \( r \) is \( \sqrt{2} \).

Step 2: Find \(\cos(\theta)\)

The cosine of angle \( \theta \) is the ratio of the x-coordinate of the point \( P \) to the radius \( r \). Therefore, \( \cos(\theta) = \frac{x}{r} \).Substitute the values:\[ \cos(\theta) = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \].

Step 3: Find \(\sin(\theta)\)

The sine of angle \( \theta \) is the ratio of the y-coordinate of the point \( P \) to the radius \( r \). Therefore, \( \sin(\theta) = \frac{y}{r} \).Substitute the values:\[ \sin(\theta) = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2} \].

Step 4: Find \(\tan(\theta)\)

The tangent of angle \( \theta \) is the ratio of the y-coordinate to the x-coordinate of the point \( P \). Therefore, \( \tan(\theta) = \frac{y}{x} \).Substitute the values:\[ \tan(\theta) = \frac{-1}{-1} = 1 \].

Key Concepts

Cosine

When we talk about cosine in trigonometry, we are dealing with the relationship between the angle and the adjacent side of a right triangle. For an angle \(\theta\), the cosine (\(\cos\theta\)) is essentially the x-coordinate of a point \(P\) on the unit circle, divided by the radius \(r\).

This gives us the formula:

• \(\cos(\theta) = \frac{x}{r}\)

In our step-by-step solution, we find that for the point \((-1, -1)\) on the terminal side of the angle \(\theta\):

• \(x = -1\)
• \(r = \sqrt{2}\)

Thus, substituting these values, we have:

• \(\cos(\theta) = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}\)

This ratio tells us how far along the x-axis the point is from the origin, proportionally speaking.

Sine

The sine function in trigonometry expresses the relationship between the angle and the opposite side of a right triangle. For an angle \(\theta\), sine (\(\sin\theta\)) is the y-coordinate of a point \(P\) on the circle divided by the radius \(r\).

The basic formula for sine is:

• \(\sin(\theta) = \frac{y}{r}\)

In the context of our problem with point \((-1, -1)\), you have:

• \(y = -1\)
• \(r = \sqrt{2}\)

Therefore, calculating the sine, we get:

• \(\sin(\theta) = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}\)

The sine function tells us how far up or down the point \(P\) is from the x-axis, in relation to the radius.

Tangent

Tangent is another fundamental trigonometric function, which is often less intuitive but equally important. It relates the sine and cosine of an angle. Specifically, for an angle \(\theta\), the tangent (\(\tan\theta\)) is the ratio of sine to cosine. In simpler terms, it's the relationship between the y and x coordinates of a point \(P\).

The formula for tangent is:

• \(\tan(\theta) = \frac{y}{x}\)

From our solution:

• \(y = -1\)
• \(x = -1\)

Thus,

• \(\tan(\theta) = \frac{-1}{-1} = 1\)

This means that for this particular angle, the change along the y-axis equals the change along the x-axis, leading to a tangent value of 1.

Radius

The radius is a crucial concept when dealing with circles in trigonometry. It is the distance from the center of the circle, usually the origin \((0,0)\), to any point \(P\) on the circle. In the context of trigonometric functions, the radius is also the hypotenuse of the right triangle formed by the x and y coordinates.

The formula to find the radius when given a point \((x, y)\) is:

• \(r = \sqrt{x^2 + y^2}\)

Applying this to our example:

• \(x = -1\)
• \(y = -1\)

So,

• \(r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}\)

Thus, the radius \(r\) here equals \(\sqrt{2}\), offering us a standardized way to assess the point's position on the circle.

Standard Position

The idea of an angle in standard position is a core concept in trigonometry and geometry. It promises a consistent way to measure angles and describe their properties. An angle is in standard position when its vertex is at the origin \((0, 0)\) of the coordinate system, and its initial side lies along the positive x-axis.

When an angle is placed in standard position:

• The terminal side represents where the angle "opens" up to.
• If a point \(P(x, y)\) is on this terminal side, then the coordinates and trigonometric ratios are uniquely defined based on that location.

For our problem, the given point \((-1, -1)\) lies on the terminal side of angle \(\theta\). This allows us to utilize trigonometric functions like sine, cosine, and tangent to describe the angle's properties using this consistent framework.