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Proving a Trigonometric Identity


Date: 06/06/99 at 21:02:00
From: Francis Lamoureux
Subject: Trigonometric identity

Two trigonometric identities I need to solve are already posted in 
your trigonometry section, yet I can't seem to figure out how to prove 
them.

Here they are, right and left side.

cos(a+b)*cos(a-b) = cos^2a - sin^2b

tan(x/2) = (1-cosx)/(sinx)

Any help on this would be greatly appreciated.


Date: 06/07/99 at 08:29:43
From: Doctor Rick
Subject: Re: Trigonometric identity

Hello, Francis.

   cos(a+b)*cos(a-b) = cos^2a - sin^2b

Use the angle sum and difference formulas to rewrite the left side:

   [cos(a)cos(b) - sin(a)sin(b)]*[cos(a)cos(b) + sin(a)sin(b)]

Expand this product using the distributive property (or FOIL) and two 
terms will cancel out. You will be left with

   cos^2(a)cos^2(b) - sin^2(a)sin^2(b)

Now use Pythagoras (in the trig form, sin^2(a) + cos^2(a) = 1) to 
replace cos^2(b) in the first term and sin^2(a) in the right term. 
Again, two terms will cancel out and you'll get the final answer.

   tan(x/2) = (1-cosx)/(sinx)

If I had to derive the right side from the left, I would have a hard 
time. Starting from the right side, it's much easier. Start by letting 
x = 2a:

   (1-cos(2a))/sin(2a)

Use the double-angle formulas; then use Pythagoras and the expression 
will reduce to tan(a), which is tan(x/2).

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

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