Deriving law of sines from cross product

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SUMMARY

This discussion focuses on deriving the Law of Sines using the cross product of vectors in a triangle defined by three vectors, ##\vec{A}##, ##\vec{B}##, and ##\vec{C}##, where ##\vec{A} + \vec{B} + \vec{C} = 0##. The key steps involve taking the cross product of these vectors and establishing relationships between the angles and the sides of the triangle. Specifically, the equations derived from the cross products lead to the relationships ##\frac{\sin(\theta_1)}{|B|} = \frac{\sin(\theta_2)}{|A|}## and ##\frac{\sin(\theta_2)}{|A|} = \frac{\sin(\theta_3)}{|C|}##, effectively tying all three angles and sides together. The discussion emphasizes the necessity of comparing two angles at a time to incorporate the third angle into the proof.

PREREQUISITES
  • Understanding of vector operations, specifically cross products.
  • Familiarity with the geometric interpretation of vectors in a triangle.
  • Knowledge of trigonometric functions, particularly the sine function.
  • Basic principles of vector addition and properties of triangles.
NEXT STEPS
  • Study the properties of vector cross products in depth.
  • Learn about the geometric interpretation of the Law of Sines.
  • Explore the derivation of the Law of Cosines as a complementary concept.
  • Investigate other vector-based proofs in trigonometry.
USEFUL FOR

This discussion is beneficial for mathematics students, educators, and anyone interested in advanced trigonometry and vector calculus, particularly those looking to understand the relationship between vectors and trigonometric laws.

Mr Davis 97
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I am trying to derive the law of signs from the cross product.

First, we have three vectors ##\vec{A} ~\vec{B} ~\vec{C}## such that ##\vec{A} + \vec{B} + \vec{C} = 0##. This creates a triangle. Then, we label the angles opposite the respective sides as a, b, and c. I am not sure where to go from here... We could take the cross product of each combination of ##\vec{A}## and ##\vec{B}##, but these cross products aren't necessarily equal, so can't set them equal to derive the law of sines... Any help would be appreciated.
 
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Mr Davis 97 said:
I am trying to derive the law of signs from the cross product.

First, we have three vectors ##\vec{A} ~\vec{B} ~\vec{C}## such that ##\vec{A} + \vec{B} + \vec{C} = 0##. This creates a triangle. Then, we label the angles opposite the respective sides as a, b, and c. I am not sure where to go from here... We could take the cross product of each combination of ##\vec{A}## and ##\vec{B}##, but these cross products aren't necessarily equal, so can't set them equal to derive the law of sines... Any help would be appreciated.
If you take ## C \times (A+B+C)=0 ## and ## C \times C=0 ## Then ## |A||C|sin(\theta_1)=|B||C|sin(\theta_2) ## (neglecting a minus sign which is of little significance).Result is ## sin(\theta_1)/|B|=sin(\theta_2)/|A| ##
 
How do I get ##sin(\theta_3)/|C|## in there? That was my main problem.
 
Mr Davis 97 said:
How do I get ##sin(\theta_3)/|C|## in there? That was my main problem.
That will show up if you take ## B \times (A+B+C)=0 ## (or ## A \times (A+B+C)=0 ##). You can only do two angles at a time by this method. Comparing two of the equations will tie the 3rd one in there. e.g. ## B \times ## gives you ## sin(\theta_2)/|A|=sin(\theta_3)/|C| ##.
 
Awesome. Thanks! While the way you put it is very understandable, but I'm not really sure how you came to the conclusion that the cross product of one of the vectors with the sum of the other three would lead you to the proof... Basically, "how would I have thought of that?"
 
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