Problem 1
Question
If the angle \(\theta\) is in standard position and \(P(x, y)\) is a point on the terminal side of \(\theta\), and \(r\) is the distance from the origin to \(P,\) then \(\sin \theta=\text{_____}\) \(\cos \theta=\text{_____}\) \(\tan \theta=\text{_____}\)
Step-by-Step Solution
Verified Answer
\(\sin \theta = \frac{y}{\sqrt{x^2+y^2}}\), \(\cos \theta = \frac{x}{\sqrt{x^2+y^2}}\), \(\tan \theta = \frac{y}{x}\).
1Step 1: Determine the Distance from the Origin to Point P
The distance from the origin (0,0) to the point \(P(x, y)\) is calculated using the Pythagorean theorem. This distance \(r\) is given by: \[ r = \sqrt{x^2 + y^2} \] where \(x\) and \(y\) are the coordinates of point \(P\).
2Step 2: Calculate the Sine of the Angle
The sine of angle \(\theta\) is given by the ratio of the opposite side to the hypotenuse in a right triangle. Thus, we have: \[ \sin \theta = \frac{y}{r} \] Substitute \(r\) from Step 1: \[ \sin \theta = \frac{y}{\sqrt{x^2 + y^2}} \]
3Step 3: Calculate the Cosine of the Angle
The cosine of angle \(\theta\) is the ratio of the adjacent side to the hypotenuse. Therefore, \[ \cos \theta = \frac{x}{r} \] Using the expression for \(r\) from Step 1: \[ \cos \theta = \frac{x}{\sqrt{x^2 + y^2}} \]
4Step 4: Calculate the Tangent of the Angle
The tangent of an angle is the ratio of the opposite side to the adjacent side. Thus, \[ \tan \theta = \frac{y}{x} \] This expression does not involve \(r\) and directly uses the coordinates of point \(P\).
Key Concepts
Pythagorean TheoremSine FunctionCosine FunctionTangent Function
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in trigonometry that comes in handy when dealing with right triangles. It asserts that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, it's represented as: \[ c^2 = a^2 + b^2 \] where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides.
In the given problem, if \( P(x, y) \) is a point on the plane, then \( r \), the distance from the origin (0,0) to the point, is treated as the hypotenuse of a right triangle. Applying the theorem here, we have: - \( x \) is one side - \( y \) is the other sideHence, the distance \( r \) becomes: \[ r = \sqrt{x^2 + y^2} \] This provides a useful way to link the geometric and trigonometric properties of the point and the angle.
In the given problem, if \( P(x, y) \) is a point on the plane, then \( r \), the distance from the origin (0,0) to the point, is treated as the hypotenuse of a right triangle. Applying the theorem here, we have: - \( x \) is one side - \( y \) is the other sideHence, the distance \( r \) becomes: \[ r = \sqrt{x^2 + y^2} \] This provides a useful way to link the geometric and trigonometric properties of the point and the angle.
Sine Function
The sine function, often abbreviated as \( \sin \), is a trigonometric function that relates the angle \( \theta \) to the ratio of the length of the opposite side to the hypotenuse in a right triangle. It is a critical function in trigonometry for understanding angle relationships. For any angle \( \theta \) in a right triangle, the sine can be defined using the formula: - \( \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} \)In our example, with the point \( P(x, y) \), the opposite side is \( y \), and the hypotenuse is \( r \). Thus, the relationship becomes: \[ \sin \theta = \frac{y}{r} \]Substituting \( r = \sqrt{x^2 + y^2} \), we get:\[ \sin \theta = \frac{y}{\sqrt{x^2 + y^2}} \]This expression helps us easily find the sine of the angle based on the coordinates of point \( P \).
Cosine Function
The cosine function, denoted as \( \cos \), is another crucial trigonometric function. It relates an angle in a right triangle to the ratio of the length of the adjacent side over the hypotenuse. The formula for cosine is: - \( \cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} \)For our scenario, if we look at the angle \( \theta \) formed by point \( P(x, y) \) in standard position, the adjacent side is \( x \), and the hypotenuse again is \( r \). Thus, cosine is calculated as:\[ \cos \theta = \frac{x}{r} \]By using the given expression for \( r \), \( \sqrt{x^2 + y^2} \), the formula transforms into:\[ \cos \theta = \frac{x}{\sqrt{x^2 + y^2}} \]This formula provides a simple way to determine the cosine of \( \theta \) using the coordinates of the point.
Tangent Function
The tangent function, indicated by \( \tan \), is a key trigonometric function often used to relate the angles and sides of a right triangle. Tangent is defined as the ratio of the length of the opposite side to the length of the adjacent side. Its basic formula is: - \( \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \)In the context of our angle \( \theta \) with point \( P(x, y) \), the formula simplifies to:\[ \tan \theta = \frac{y}{x} \]Here, you notice that \( r \) does not appear in the calculation of tangent. This simplification shows that tangent is entirely dependent on the coordinates \( x \) and \( y \) of point \( P \), making it a straightforward calculation in many cases. Understanding this relationship helps in applying tangent to solve various trigonometric problems.
Other exercises in this chapter
Problem 1
For triangle \(A B C\) with sides \(a, b,\) and \(c\) the Law of Cosines states \(c^{2}=\) ________________________
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The inverse sine, inverse cosine, and inverse tangent functions have the followings domains and ranges. (a) The function \(\sin ^{-1}\) has domain _______ and r
View solution Problem 2
In which of the following cases must the Law of cosines be used to solve a triangle? $$ASA\quad SSS\quad SAS\quad SSA$$
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In which of the following cases can we use the Law of Sines to solve a triangle? ASA SSS SAS SSA
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