cos at cos B² + (sina tsin B 2-4c0? (2-B
Answers
Answer:
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).