High School Derivation of Cosine and Sine Method of Vector Sum

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SUMMARY

The discussion focuses on the derivation of the cosine and sine rules for vector addition, highlighting their application in physics. The cosine rule is expressed as $$\vec {R^2} = \vec {F^2_1} + \vec {F^2_2} + 2 \cdot \vec F_1 \cdot \vec F_2 \cdot cos ~ \alpha$$, while the sine rule is given by $$\frac {\vec R} {sin ~ \alpha} = \frac {\vec F_1} {\sin ~ (\alpha - \beta)} = \frac {\vec F_2} {\sin ~ \beta}$$. The discussion contrasts these methods with the component-based approach commonly taught in undergraduate physics, emphasizing that both methods are valid and depend on the context of the problem.

PREREQUISITES
  • Understanding of vector addition and its geometric representation
  • Familiarity with the cosine and sine rules in trigonometry
  • Basic knowledge of scalar and vector products
  • Ability to interpret angles and sides in triangles
NEXT STEPS
  • Study the derivation of the cosine rule in detail
  • Explore the applications of the sine rule in various physics problems
  • Learn about vector components and their role in physics
  • Investigate the geometric interpretation of vector addition
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Students of physics, educators teaching vector mechanics, and anyone interested in the mathematical foundations of vector addition methods.

bagasme
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Hello all,

In high school physics, the magnitude sum of vector addition can be found by cosine rule:

$$\vec {R^2} = \vec {F^2_1} + \vec {F^2_2} + 2 \cdot \vec F_1 \cdot \vec F_2 \cdot cos ~ \alpha$$

and its angle are calculated by sine rule:

$$\frac {\vec R} {sin ~ \alpha} = \frac {\vec F_1} {\sin ~ (\alpha - \beta)} = \frac {\vec F_2} {\sin ~ \beta}$$

where ##\alpha## is the angle between two vectors, and ##\beta## is the angle between ##\vec F_1## and ##\vec R##.

In undergraduate physics books, however, the methods above are not taught, instead the vector addition is done by components, and use ##tan ~ \theta = \frac {R_y} {R_x}## to obtain the angle.

How are derivations of vector addition formulas by cosine and sine rule (as above), and in what cases those formulas can be used in place of vector addition by components?

Bagas
 
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Draw the vectors as arrows. Place the tail of one arrow at the tip of the other. This forms two sides of a triangle. The vector sum is the third side of the triangle, with its head at the head of the second vector.

You now have a triangle with two known sides and one known angle. Any method you like to find the length and angle of the third side is acceptable. Looks like your high school textbook preferred classical geometry. A lot of physicists almost always just resolve the vectors into components and prefer that approach. Neither approach is wrong - its just personal preference and judgement based on exactly what knowns and unknowns you have.
 
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bagasme said:
In undergraduate physics books, however, the methods above are not taught ...

You'll always need the rule of cosines and the rule of sines. They are generally useful at all levels of physics.
 
Note that first of all one has to guess the meaning of your symbols and 2nd most probably the formulae you quote are wrong.

I guess you want to show the cosine and sine theorems for triangles. Consider an arbitrary triangle with points ##A##, ##B##, and ##C##.

The cosine rule is most simple to derive. For that you only need
$$\overrightarrow{AB} + \overrightarrow{BC}=\overrightarrow{AC} \; \Rightarrow \; \overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}.$$
Taking the square in the sense of the scalar product of this yields
$$|BC|^2=|AC|^2 + |AB|^2 -2 \overrightarrow{AC} \cdot \overrightarrow {AB}=|AC|^2 + |AB|^2 - 2 |AC| |AB| \cos \alpha,$$
where ##\alpha## is the angle at the point ##A##. This is the cosine rule.

The sine rule is most easily derived by calculating the area of the triangle with help of the cross product. You can use any two of the vectors making up the triangle you like:
$$2F=|\overrightarrow{AB} \times \overrightarrow{AC}|=|AB| |AC| \sin \alpha$$
or
$$2F=|\overrightarrow{BA} \times \overrightarrow{BC}|=|AB| |BC| \sin \beta$$
Since both are equal you get
$$|AC| \sin \alpha =|BC| \sin \beta \; \Rightarrow \; \frac{|AC|}{\sin \beta} = \frac{BC}{\sin \alpha},$$
which is the sine rule.
 
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bagasme said:
In undergraduate physics books, however, the methods above are not taught

Neither is counting on your fingers and toes.
 
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