Question 1204374: If sin 𝛼 + sin 𝛽 = 𝑝, cos 𝛼 − cos 𝛽 = 𝑞, where 𝑝 ≠ 0 and 𝑝 ≠ 𝑞, find cot(𝛼 − 𝛽) in terms of
𝑝 and 𝑞.
Answer by math_tutor2020(3817)   (Show Source):
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Answer: (q^2 - p^2)/(2pq)
Explanation
I'll use A in place of alpha, and B in place of beta.
Refer to page 2 of this trig identity sheet
https://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf
Similar reference sheets should be found in the back of your trig textbook, and this type of sheet should be given to you by your professor when exams happen.
Look at the "Sum to Product Formulas" section to say the following:
p = sin(A) + sin(B) = 2*sin( (A+B)/2 )*cos( (A-B)/2 )
q = cos(A) - cos(B) = -2*sin( (A+B)/2 )*sin( (A-B)/2 )
Then divide q over p
The 2's out front cancel. So do the sin( (A+B)/2) terms.
We'll end up with something of the form sin/cos, which represents tangent.
We get
q/p = -tan( (A-B)/2 )
which rearranges to
tan( (A-B)/2 ) = -q/p
Then notice
tan(A-B) = tan(2*(A-B)/2) = tan(2*C) where C = (A-B)/2
which furthermore means
tan( (A-B)/2 ) = tan(C) = -q/p
We do this so we can use the tangent identity located in the "Double Angle Formulas" section
tan(2*C) = 2*tan(C)/(1 - tan^2(C))
tan(2*C) = 2*(-q/p)/(1 - (-q/p)^2)
tan(2*C) = (-2q)/(p(1 - (-q/p)^2))
tan(2*C) = (-2q)/(p - (q^2)/p)
tan(2*C) = (-2pq)/(p^2 - q^2)
tan(A-B) = (-2pq)/(p^2 - q^2)
cot(A-B) = -(p^2 - q^2)/(2pq)
cot(A-B) = (q^2 - p^2)/(2pq)
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