There is an arbitrary triangle with angles A, B, and C and sides of lengths a, b, and c. Angle A is opposite side a.
How do I get the formulas:
b * cos C + c * cos B = a
c * cos A + a * cos C = b
a * cos B + b * cos A = c
Are these standard trig formulas? What are they called? Or where are they derived from?
thanks!
b * cos C + c * cos B = a
Draw the line at right angles to side a that splits the angle A (not necessarily in equal parts). Then, the triangle is slit in two right triangles, and you see that
a = a1 + a2
And:
a1 = b * cos C
a2 = c * cos B
Cyclically permuting a -- > b---> c ---> a (and the uppercase variables) give you the other two equations)
1-2sin^5B write in single trigonometric function
These formulas are derived from the Law of Cosines, which relates the lengths of the sides of a triangle to the cosines of its angles. The Law of Cosines states that in any triangle with side lengths a, b, and c and corresponding opposite angles A, B, and C:
a^2 = b^2 + c^2 - 2bc * cos(A)
b^2 = c^2 + a^2 - 2ca * cos(B)
c^2 = a^2 + b^2 - 2ab * cos(C)
To derive the formulas you mentioned, we can rearrange the equations above:
Starting with a^2 = b^2 + c^2 - 2bc * cos(A), we can isolate the cos(A) term:
2bc * cos(A) = b^2 + c^2 - a^2
cos(A) = (b^2 + c^2 - a^2) / (2bc)
Multiplying both sides by a gives:
a * cos(A) = (a * (b^2 + c^2 - a^2)) / (2bc)
a * cos(A) = (ab^2 + ac^2 - a^3) / (2bc)
By cyclically permuting the variables in the equation above, we obtain your three formulas:
b * cos(C) + c * cos(B) = a
c * cos(A) + a * cos(C) = b
a * cos(B) + b * cos(A) = c
These formulas can be helpful in solving for unknown sides or angles in a triangle when you know the lengths of two sides and the measure of one angle. They are often used in trigonometry and geometry.