Proving a Trigonometric IdentityDate: 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/ |
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