Problem 18
Question
Find the exact values of the trigonometric functions for the acute angle \(\theta\). $$\cos \theta=\frac{8}{17}$$
Step-by-Step Solution
Verified Answer
\( \sin \theta = \frac{15}{17} \), \( \tan \theta = \frac{15}{8} \).
1Step 1: Identify the Trigonometric Relationship
Given that \( \cos \theta = \frac{8}{17} \), use the identity \( \cos \theta = \frac{adjacent}{hypotenuse} \). This indicates that the side adjacent to angle \( \theta \) is 8 units long, and the hypotenuse is 17 units long.
2Step 2: Apply the Pythagorean Theorem
To find the opposite side, use the Pythagorean theorem: \( a^2 + b^2 = c^2 \),where \( a \) is the adjacent side, \( b \) is the opposite side, and \( c \) is the hypotenuse. Substituting the known values, we have:\[ 8^2 + b^2 = 17^2 \]\[ 64 + b^2 = 289 \]\[ b^2 = 225 \]\[ b = 15 \].Thus, the opposite side is 15 units long.
3Step 3: Calculate \( \sin \theta \)
The sine of an angle is the quotient of the length of the opposite side and the hypotenuse: \( \sin \theta = \frac{opposite}{hypotenuse} \). Using the values calculated:\[ \sin \theta = \frac{15}{17} \].
4Step 4: Calculate \( \tan \theta \)
The tangent of an angle \( \theta \) is given by the ratio of the opposite to the adjacent side: \( \tan \theta = \frac{opposite}{adjacent} \). Using the values found:\[ \tan \theta = \frac{15}{8} \].
5Step 5: Verify Using Trigonometric Identities
Verify the values using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):\[ \left( \frac{15}{17} \right)^2 + \left( \frac{8}{17} \right)^2 = 1 \]\[ \frac{225}{289} + \frac{64}{289} = \frac{289}{289} = 1 \].The result holds; thus, the calculations are verified.
Key Concepts
Pythagorean TheoremSine FunctionCosine FunctionTangent Function
Pythagorean Theorem
The Pythagorean Theorem is a fundamental principle in geometry that establishes the relationship between the sides of a right-angled triangle. This theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. Mathematically, it's expressed as:
Using this theorem helps us find any missing side of a right triangle if two side lengths are known. In our exercise, knowing
- \( a^2 + b^2 = c^2 \)
Using this theorem helps us find any missing side of a right triangle if two side lengths are known. In our exercise, knowing
The Role in the Problem:
Given: \( \cos \theta = \frac{8}{17} \). The adjacent side is 8, and the hypotenuse is 17. To find the opposite side, apply:- \( 8^2 + b^2 = 17^2 \)
- Solve: \( 64 + b^2 = 289 \)
- Leads to: \( b^2 = 225 \)
- So, \( b = 15 \)
Sine Function
The sine function is fundamental in trigonometry. For a right-angled triangle, the sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. This is expressed as:
- \( \sin \theta = \frac{opposite}{hypotenuse} \)
Applying Sine to the Problem:
With the opposite side found to be 15 and hypotenuse as 17:- \( \sin \theta = \frac{15}{17} \)
Cosine Function
The cosine function measures the ratio between the adjacent side and the hypotenuse in a right triangle. It's defined by the expression:
- \( \cos \theta = \frac{adjacent}{hypotenuse} \)
Utilizing Cosine in Our Exercise:
We were given \( \cos \theta = \frac{8}{17} \) directly in our initial problem:- The adjacent side was 8, and the hypotenuse was 17.
- Had this been unknown, we could find these values by rearranging and solving the formula \( \cos \theta = \frac{8}{17} \).
Tangent Function
The tangent function in trigonometry is the ratio of the length of the opposite side to that of the adjacent side in a right triangle. This is expressed with the formula:
- \( \tan \theta = \frac{opposite}{adjacent} \)
Finding Tangent in Our Solution:
Using the values already found, the tangent function was calculated as:- \( \tan \theta = \frac{15}{8} \)
Other exercises in this chapter
Problem 18
Find the period and sketch the graph of the equation. Show the asymptotes. $$ y=-3 \tan \left(\frac{1}{3} x-\frac{\pi}{3}\right) $$
View solution Problem 18
Find the exact value. (a) \(\csc (3 \pi / 4)\) (b) \(\csc (-2 \pi / 3)\)
View solution Problem 18
Express \(\theta\) in terms of degrees, minutes, and seconds, to the nearest second. $$\theta=1.5$$
View solution Problem 19
Exer. 17-24: Given the indicated parts of triangle \(A B C\) with \(\gamma=90^{\circ}\), express the third part in terms of the first two. $$ \beta, b ; \quad a
View solution