Question

cos at cos B² + (sina tsin B 2-4c0? (2-B​

Answer 1

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).