How to Use Derivation of the Law of Cosines
TL;DR: Wondering where the Law of Cosines actually comes from? Drop an altitude inside a general triangle, splitting it into two right triangles, then apply the Pythagorean theorem twice and do a bit of algebra cleanup. Out pops c squared equals a squared plus b squared minus 2ab times cosine of C. The point isn't just to admire the proof, once you've seen the derivation, the formula stops being random memorization and starts feeling like an old friend you actually understand.
Key takeaways:
- Place one vertex at the origin with one side along the x-axis.
- Drop an altitude from the third vertex to form two right triangles.
- Apply the Pythagorean theorem to the right triangle on the far side.
- The result is \(c^2 = a^2 + b^2 - 2ab\cos C\).
- The same approach works in vector form using the dot product.
The Law of Cosines, an invaluable tool in trigonometry, opens up plenty of problem-solving avenues in mathematics and physics. The process of deriving this law is a critical step to fully understand its potential. This guide aims to elucidate the derivation of the Law of Cosines and how to employ it, intertwined with complex concepts and contextual applications.
Step-by-step Guide to How to Use Derivation of the Law of Cosines
Here is a step-by-step guide to how to use a derivation of the law of cosines:
Type the problems from this lesson and check each step.
Step 1: Comprehending the Law of Cosines
Start by firmly understanding the Law of Cosines. It states that for any triangle (not just right-angled) with sides of lengths \(a, b\), and \(c\), and the angle \(γ\) (gamma) opposite side \(c\), the following relationship holds: \(c² = a² + b²\ – 2ab\times cos(γ)\). Get to know this formula well, as it forms the backbone of the entire derivation process.
Step 2: Familiarizing Yourself with Vector Operations
Next, acquaint yourself with the concept of vectors. Understand their properties, and learn how to perform vector operations like addition, subtraction, and multiplication (both dot product and cross product). Grasp how to express a vector in terms of its components.
Step 3: Learning the Geometric Interpretation of Dot Product
The geometric interpretation of the dot product (or scalar product) of two vectors is the foundation of the Law of Cosines. When two vectors are dot-multiplied, the result is equal to the product of the vectors’ magnitudes and the cosine of the angle between them. Learn and appreciate this essential principle.
Step 4: Deriving the Law of Cosines
Now, walk through the derivation itself. Consider an arbitrary triangle with vertices \(A, B\), and \(C\), and denote the vectors \(AB\) and \(AC\) by \(a\) and \(b\), respectively. The third side, \(BC\), can then be expressed as the vector \(c = b\ – a\). Square the magnitude of \(c\) and equate it with the dot product interpretation. This approach will lead you to the Law of Cosines.
Step 5: Comprehending the Derivation
After following the derivation, take a moment to fully understand it. Recognize how the formula was derived using vector principles and observe that it holds for any triangle, not just right-angled ones. This step is vital for deeply comprehending the Law of Cosines and its applications.
Step 6: Exploring Applications of the Law of Cosines
Once you understand the derivation, you can explore how to apply the Law of Cosines in different scenarios. It has extensive applications in diverse fields such as physics, engineering, computer science, and navigation. The law is particularly useful in calculating unknown side lengths or angles in a triangle when certain other measures are known.
Step 7: Applying the Law of Cosines in Problem-Solving
Put your understanding to the test by using the Law of Cosines to solve problems. Begin with simple problems, like finding the length of a side or an angle in a triangle. Gradually, take on more complex problems, such as calculating the distance between two points in three-dimensional space, or solving real-life navigation problems.
By following this step-by-step guide, you’ll gain a thorough understanding of the derivation and use of the Law of Cosines. This knowledge will prove indispensable in a variety of mathematical and scientific endeavors, showing the elegant power of trigonometry and vector calculus.
Recommended EffortlessMath Books
For a complete trig workbook that walks through every derivation alongside the applications, the Trigonometry for Beginners (resource currently unavailable) covers law of cosines, law of sines, and the trig identities with worked examples. For precalc-level prep that ties trig to vectors and analytic geometry, see Pre-Calculus for Beginners.
Frequently Asked Questions
What is the law of cosines?
The law of cosines links all three sides of a triangle with one of its angles: c² = a² + b² − 2ab cos C, where C is the angle opposite side c. It works for any triangle. When C = 90°, cos C = 0 and the formula becomes c² = a² + b², so the Pythagorean theorem is really a special case of it.
How do you derive the law of cosines?
One short derivation uses vectors. Let the two sides meeting at angle C be vectors a and b; the third side is then a − b. Squaring its length gives |a − b|² = |a|² + |b|² − 2(a · b). Because the dot product a · b equals ab cos C, this becomes c² = a² + b² − 2ab cos C. The minus sign comes straight from the cross term in that expansion.
How do you derive the law of cosines using the Pythagorean theorem?
Drop an altitude h from vertex B to side AC, meeting it at point D. Then CD = a cos C, AD = b − a cos C and h = a sin C. The Pythagorean theorem in right triangle ABD gives c² = (b − a cos C)² + (a sin C)². Expand, replace sin²C + cos²C with 1, and the result is c² = a² + b² − 2ab cos C.
How do you use the law of cosines?
With two sides and the angle between them (SAS), plug straight into c² = a² + b² − 2ab cos C and take the square root. For sides 6 and 9 with a 120° angle between them, cos 120° = −0.5, so c² = 36 + 81 + 54 = 171 and c ≈ 13.1. A negative cosine makes the third side longer than Pythagoras alone would give, which fits the wide angle.
How do you use the law of cosines with SSS?
When all three sides are known, solve the formula for the cosine: cos C = (a² + b² − c²) ÷ (2ab). Take sides 7, 8 and 10. The angle opposite 10 has cos C = (49 + 64 − 100) ÷ 112 = 13/112 ≈ 0.116, so C ≈ 83.3°. Find the largest angle first; if its cosine is negative, the triangle is obtuse.
How do you use the law of cosines and sines together?
Start with the law of cosines for SAS or SSS, since the law of sines needs a known angle and its opposite side. Once you have that pair, switch to the law of sines. Example: a = 5, b = 8, C = 60° gives c² = 25 + 64 − 40 = 49, so c = 7. Then sin A = 5 × sin 60° ÷ 7 ≈ 0.619, so A ≈ 38.2° and B ≈ 81.8°.
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