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Law of Tangents


Date: 01/22/99 at 01:23:14
From: Matthew Veatch
Subject: Law of tangents

For my pre-calc class, I am trying to prove and derive the law of 
tangents. Please help me. Thank you.


Date: 01/22/99 at 12:44:51
From: Doctor Floor
Subject: Re: Law of tangents

Hi Matthew,

Thank you for sending your question to Dr. Math!

Let a and b be two sides of a triangle. Let A be the angle opposite a 
and B be the angle opposite b. The law of tangents says:

  a+b   tan(0.5(A+B))
  --- = -------------
  a-b   tan(0.5(A-B))

To derive this formula, first consider the sum and subtraction formulas 
for sines:

  sin(t+u) = sin(t)cos(u) + cos(t)sin(u)
  sin(t-u) = sin(t)cos(u) - cos(t)sin(u)

Adding and subtracting these two gives:

  sin(t+u) + sin(t-u) = 2sin(t)cos(u)
  sin(t+u) - sin(t-u) = 2cos(t)sin(u)

Now let t = 0.5(A+B) and u = 0.5(A-B), then t+u = A and t-u = B. 
This gives:

  sin(A) + sin(B) = 2sin(0.5(A+B))cos(0.5(A-B))
  sin(A) - sin(B) = 2sin(0.5(A-B))cos(0.5(A+B))

So we can derive the following:

  tan(0.5(A+B))   sin(0.5(A+B))cos(0.5(A-B))
  ------------- = --------------------------
  tan(0.5(A-B))   sin(0.5(A-B))cos(0.5(A+B))

                  2sin(0.5(A+B))cos(0.5(A-B))
                = ---------------------------
                  2sin(0.5(A-B))cos(0.5(A+B))

                  sin(A) + sin(B)
                = ---------------
                  sin(A) - sin(B)

And using the law of sines this equals:

                   a + b
                =  -----
                   a - b

as desired!

If you have a math question again, please send it to Dr. Math.

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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